When solving problems involving algebraic identities it is crucial to identify them and based on this identification apply correct algebraic fromula. There are 8 basic indetities and these are:
- the sqaure of binomial can be described as the sum of: the square of the first term, twice the product of the two terms and the suqare of the last term. $$(x+y)^2 = x^2 + 2xy + y^2 $$ $$ (x-y)^2 = x^2 - 2xy + y^2$$
- the cube of a binomial - there are two types of the cube of the binomial and these are sum of cubes and the difference of cubes
- the sum of cubes - is the sum of the cube of the first term, plus three times the square of the first term by the second terms, plus three times the first term follwed by the square of the second term and finally the cube of the second term. The sum of cubes can be written in mathematical form as: $$(a+b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3$$
- the difference of cubes - is the differece of cubes of two binomials is the cube of the first term, minus three times the square of the first term follwed by second term, plus three times the first term follwed by the square of the second term and finally minus the cube of the second term. The difference of the cubes can be written in mathematica form as: $$(a-b)^3 = a^3 - 3a^2 b + 3ab^2 - b^3$$
- the difference of the squares - the difference of two squares is multiplcation of the difference between first and second term and the addition of first and second term. Mathematically this relation is written as: $$ a^2 - b^2 = (a-b)(a+b)$$ The previous relation is thrue and here is the proof. \begin{eqnarray}\require{cancel}(a-b)(a+b) &=& a^2 + \cancel{ab} - \cancel{ab} - b^2 \\ (a-b)(a+b) &=& a^2-b^2 \end{eqnarray}
- the sum of cubes - the sum of cubes \(a^3 + b^3\) can be written as the multiplication of two parenthesis. In first parenthesis the sum of first and second term are located while in the second parenthesis the square of first term, minus first term multipled by the second term, plus the square of the second term are located. The sum of cubes can be written as: $$ a^3 + b^3 = (a+b)(a^2-ab +b^2) $$ The proof of previous relation can be written as \begin{eqnarray}\require{cancel} (a+b)(a^2-ab+b^2) &=& a^3 - \cancel{a^2b} + \cancel{ab^2} + \cancel{a^2b} -\cancel{ab^2} + b^3\\ (a+b)(a^2-ab+b^2) &=& a^3 +b^3 \end{eqnarray}
- the difference of cubes - the difference of cubes \(a^3 - b^3\) just like the sum of cubes can be written as the mulitplication of two parenthesis. In the first parenthesis the difference between first and second term are givne while in the second parenthesis the square of the first member , plus the first term multiplied by the second and finally the square of the second term is given. This description of the difference of cubes in mathematical form can be written as: $$ a^3 - b^3 = (a-b)(a^2 +ab +b^2)$$ The proof of previous relation is given below. \begin{eqnarray}\require{cancel}(a-b)(a^2 + ab+ b^2) &=& a^3 + \cancel{a^2b} + \cancel{ab^2} -\cancel{a^2b} - \cancel{ab^2} - b^3\\ (a-b)(a^2 + ab+ b^2) &=& a^3 - b^3\end{eqnarray}
- \((a-b)^2 = (b-a)^2\)
- \((a-b) = -(b-a)\)
- \(-a-b = -(a+b)\)
Algebraic formulas/identities - Examples
- Example 1 Calculate \(\left(\frac{1}{4}a + 2\cdot b^3\right)^2\)
Solution $$ \left(\frac{1}{4}a + 2\cdot b^3\right)^2 = \left(\frac{1}{4}\cdot a\right)^2 + 2 \cdot \frac{1}{4}\cdot a \cdot 2 \cdot b^3 + \left(2\cdot b^3\right)^2 = \frac{a^2}{16} + a\cdot b^3 + 4\cdot b^6.$$ - Example 2 Calculate \(\left(\frac{3}{5}a^2b + \frac{5}{3}ab^2\right)^3\)
Solution $$ \left(\frac{3}{5}a^2b + \frac{5}{3}ab^2\right)^3 = \left(\frac{3}{5}a^2b\right)^3+3\left(\frac{3}{5}a^2b\right)^2\left(\frac{5}{3}ab^2\right) + 3\left(\frac{3}{5}a^2b\right)\left(\frac{5}{3}ab^2\right)^2 + \left(\frac{5}{3}ab^2\right)^3 = $$ $$ = \frac{27}{125}a^6b^3 + \frac{9}{5}a^5b^4 + 5a^4b^5 + \frac{125}{27}a^3b^6$$ - Example 3 Calculate \((4-x)(3+x)-1+(x-3)^2\)
Solution $$(4-x)(3+x)-1+(x-3)^2 = 12 +4x-3x-x^2 -1 + x^2 -6x +9 = 20-5x = 5(4-x)$$ - Example 4 Calculate \(\left(\frac{1}{2} - a\right)\left(\frac{1}{2} + a\right)\)
Solution $$\left(\frac{1}{2} - a\right)\left(\frac{1}{2} + a\right) = \frac{1}{4} + \frac{1}{2} a - \frac{1}{2}a - a^2$$ - Example 5 Calculate \((n + (n+1)+(n+2))^3\)
Solution $$(n + (n+1)+(n+2))^3 = (n+n+1+n+2)^3 = (3n+3)^3 = 27n^3 + 3\cdot (3n)^2 \cdot 3+ 3\cdot 3n \cdot 3^2 + 3^3 = 27n^3 + 81n^2 + 81n + 27$$ - Example 6 Calcualte \((a-2)(a+2)(a^2+4)(a^4+16)\)
Solution$$ (a-2)(a+2)(a^2+4)(a^4+16) = (a^2-4)(a^2+4)(a^4+16) = (a^4-16)(a^4+16) = a^8 -256.$$ - Example 7 The expression \((3m-2)^2\) is equal to
- \(3m^2-6m+2\)
- \(9m^2-6m+4\)
- \(9m^2-12m+4\)
- \(3m^2 - 12m +2\)
Solution$$ (3m-2)^2 = 9m^2 -2\cdot 3m \cdot 2 + 4 = 9m^2 - 12m + 4$$ The answer is \((3m-2)^2 = 9m^2-12m+4\) which is the third solution. - Example 8 If \(x^2-y^2 = 75\) and \(x+y=15\)
calculate \(x-y\) and \(2x-2y+1\)
SolutionLet's start from second equation and express the variable \(x\) over the variable \(y\) $$x = 15-y$$ Insert that solution into the frist equation. $$ \begin{eqnarray}x^2 -y^2 &=& 75 \\ \nonumber (15-y)^2-y^2 &=&75 \\ \nonumber 225 - 30y + y^2-y^2 &=&75 \\ \nonumber -30y &=& -150 \\\nonumber y &=& 5\end{eqnarray}$$ Now that we have the determine the value of the \(y\) variable insert it back to the expression \(x = 15-y\) to obtain the value of the \(x\) variable. $$ x = 15 - y = 15 - 5 = 10$$ Now that we have determine the values of \(x\) and \(y\) variable the values of expressions \(x-y\) and \(2x-2y +1\) can be determined. $$ x-y = 10 -5 = 5$$ $$ 2x - 2y + 1 = 2\cdot 10 - 2 \cdot 5 +1 = 20 - 10 +1 = 11 $$ - Example 9 Calculate the expression \(\left(2x-3\right)^2\)
Solution $$ \left(2x-3\right)^2 = 4x^2 -2 \cdot 2x \cdot 3 + 9 = 4x^2 -12x + 9$$ - Example 10 Calculate the expression \(\left(3+2x\right)^2\)
Solution $$\left(3+2x\right)^2 = 9 + 2\cdot 3 \cdot 2x + 4x^2 = 9+12x+4x^2 $$ - Example 11 If \(x+2y =11\), determine how much is it \(x^2+4xy + 4y^2 +7\)
Solution The second equation must be simplified and after it is simplifed the first equation can be inserted. $$x^2 +4xy+4y^2 + 7 = \left(x+2y\right)^2 + 7 = 11^2 +7 = 128$$ - Example 12 Simplify the equation \((x-4)(x-3).\)
Solution $$ (x-4)(x-3) = x^2-3x -4x + 12 = x^2 -7x + 12 $$ - Example 13 Write the expression \(\left(64-x^2\right)\) in the form of the product
Solution $$ 64-x^2 = (8-x)(8+x)$$ - Example 14 Write the expression \(27x^3 - 64\) in the form of the product
Solution To solve this problem the rule \((a\pm b)(a^2 \pm ab + b^2) = a^3 \pm b^3\) called sum or difference of cubes, must be applied. $$ 27x^3 - 64 = (3x - 4)(3x^2 + 12x + 16). $$ - Example 15 Write the expression \(54a^3 - 81ab^2\) in the form of the product
Solution First the exclusion of a common factor rule must be applied and then the difference of squares rule \((a^2 -b^2) = (a+b)(a-b)\) must be applied. $$ 54a^3 - 81ab^2 = 27a(2a^2-3b^2) = 27a(\sqrt{2}a-\sqrt{3}b)(\sqrt{2}+\sqrt{3}b)$$ - Example 16 Write the expression \(25a^2 - 2ab + \frac{1}{25}b^2\) in the form of the product
Solution To transform this expression in form of the product the square binomial rule must be applied i.e. \(a^2 \pm 2ab + b^2 = (a\pm b)^2.\) $$ 25a^2 - 2ab + \frac{1}{25}b^2 = \left(5a-\frac{1}{5}b\right)^2.$$ - Example 17 Write the expression \(16a^2 - 8ab - 15b^2\) in the form of the product
SolutionTo write this expression in form of the product first the third member of the expression \(-15b^2\) must be expanded as \(b^2 - 16b^2\). The expression can be written in the following form $$ 16a^2 - 8ab - 15b^2 = 16a^2 - 8ab +b^2 - 15b^2$$ Then the square binomial rule must be applied. The expression can be written in the following form $$ 16a^2 - 8ab + b^2 -15b^2 = (4a - b)^2 - (4b)^2$$ Finnaly the difference of squares rule can be applied to obtain the expression in the form of the product. $$ (4a - b)^2 - (4b)^2 = (4a-b-4b)(4a-b+4b) = (4a-5b)(4a+3b)$$ - Example 18 Write the expression \(63ab^2 - 84a^2b + 28a^3\) in the form of the product.
SolutionFrist the extraction of the common factor must be applied. In this case the common factor is \(7a\). $$ 63ab^2 - 84a^2b + 28a^3 = 7a(9b^2 - 12ab + 4a^2).$$ Then the square binomial rule is applied. $$ 7a(9b^2 - 12ab + 4a^2) = 7a(3b-2a)^2$$ - Example 19 Write the expression \(6a^2 - 5a - 6\) in the form of the product
SolutionFirst the term \(-5a\) must be expanded and written as \(4a - 9a. \) $$6a^2 -5a -6 = 6a^2 +4a - 9a-6$$ Then the common factor \(2a\) can be extracted from first two members \(2a(3a +2)\) while the common factor of \(-3\) can be extracted from last two members. After that the third common factor \((3a+2)\) can be extracted $$ 6a^2 + 4a - 9a -6 = 2a(3a+2) - 3(3a-2) = (2a-3)(2a+3)$$ - Example 20 Write the expression \(27a^3 + 9a^2b^2 + ab^4 + \frac{b^6}{27}\) in the form of the product.
Solution In this case the cube binomial rule \(\left(\left(a\pm b\right)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3\right)\) is applied. $$ 27a^3 + 9a^2b^2 + ab^4 + \frac{b^6}{27} = \left(3a + \frac{1}{3}b^2\right)^3 $$ - Example 21 Write the expression \(4a^2 -4a-9b^2 +1\) in the form of the product.
Solution$$ 4a^2 -4a -9b^2 +1 = (2a-1)^2 - (3b)^2 = (2a-1-3b)(2a-1+3b).$$ - Example 22 Write the expression \(45a^2c - 20b^2c + 54a^2d - 24b^2d\) in the form of the product
Solution$$ 45a^2c - 20b^2c + 54a^2d - 24b^2d = 5c(9a^2 - 4b^2) + 6d(9a^2 -4b^2) = (9a^2 -4b^2)(5c+6d) = (3a-2b)(3a+2b)(5c+6d)$$ - Example 23 Write the expression \(6a^2bc + 9ab^2c^2 + 18ac^2 + 12bc\) in the form of the product
Solution $$ 6a^2bc + 9ab^2c^2 + 18ac^2 + 12bc = 3c(2a^2b+3ab^2c + 6ac + 4b)$$ - Example 24 Write the expression \(27a^4 + 54a^3 -8a - 16\) in the form of the product
Solution$$ 27a^4 + 54a^3 -8a - 16 = 27a^3(a+2) -8(a+2) = (a+2)(27a^3-8) = (a+2)(3a-2)(9a^2+6a+4).$$ - Example 25 Write the expression \(50c^4d - 60c^3d^3+18c^2d^5\) in the form of the product
Solution$$ 50c^4d - 60c^3d^3+18c^2d^5 = 2c^2d(25c^2-30cd^2+9d^4) = 2c^2d\left(5c-3d^2\right)^2$$ - Example 26 Write the expression \(36a^4-\left(9a^2 +b^2\right)^2\) in the form of the product
Solution$$ 36a^4-\left(9a^2 +b^2\right)^2 = \left(6a^2 - \left(9a^2+b^2\right)\right)\left(6a^2 + \left(9a^2+b^2\right)\right)$$ $$ = \left(6a^2 - 9a^2-b^2\right)\left(6a^2 + 9a^2+b^2\right) = \left(-3a^2-b^2\right)\left(15a^2+b^2\right).$$ - Example 27 Write the expression \(2x^4y-x^2y^2 -x^6\) in the form of the product.
Solution $$2x^4y-x^2y^2-x^6 = -x^2\left(x^4-2x^2y+y^2\right) = -x^2\left(x^2-y\right)^2 $$ - Example 28 If \(a+b=5\) and \(ab=-3\) determine how much is \(a^2-b^2.\)
Solution$$\begin{eqnarray} (a+b)^2 &=& a^2+2ab+b^2\\ \nonumber (a+b)^2-2ab &=& a^2 + b^2\\ \nonumber a^2+b^2 &=& (a+b)^2-2ab \\ \nonumber a^2+b^2 &=&5^2-2\cdot(-3) \\ \nonumber a^2 +b^2 &=& 25+6 \\ \nonumber a^2 + b^2 &=& 31\end{eqnarray} $$ - Example 29 If \(\frac{1}{4}x^2 +\frac{1}{9}y^2 -2x -2y + 13 = 0\) in the form of the product
Solution$$ \begin{eqnarray}\frac{1}{4}x^2 +\frac{1}{9}y^2 -2x -2y + 13 &=& 0 \\\nonumber \frac{1}{4}x^2-2x+4 +\frac{1}{9}y^2 -2y + 9 &=& 0 \\ \nonumber \left(\frac{1}{2}x-2\right)^2 +\left(\frac{1}{3}y -3\right)^2 &=& 0 \\\nonumber \frac{1}{2}x -2 = 0 \quad \frac{1}{3}y-3 &=& 0 \\ \nonumber x = 4 \quad y&=&9\end{eqnarray}$$ - Example 30 Write the expression \((a^3 + a^2b+ab^2 +b^3)(a-b)\) in the form of the product
Solution$$ (a^3 + a^2b+ab^2 +b^3)(a-b) = a^4-b^4$$ - Example 31 Calculate \(\left(2a^3 - \frac{1}{4}b\right)^2\)
Solution$$ \left(2a^3 - \frac{1}{4}b\right)^2 = 4a^6 - 2a^3\cdot \frac{b}{4} + \frac{b^2}{16} = 4a^6 - \frac{1}{4}a^3b + \frac{1}{16}b^3$$ - Example 32 Calculate \(\left(3a^4b^2 + 4c^3\right)^3\) in the form of the product.
Solution $$ \left(3a^4b^2 + 4c^3\right)^3 = 27a^12b^6 + 3\cdot 9a^8 b^4 \cdot 4 c^3 + 3 \cdot 3a^4b^2\cdot 16c^6 + 64c^9 = 27a^12b^6 + 108a^8b^4c^3 + 144 a^4b^2c^6 + 64c^9$$ - Example 33 Calculate \(1+(x-3)^2 - (x-4)(3+x)\)
Solution$$ 1+(x-3)^2 - (x-4)(3+x) = 1+(x^2 +6x+9) - (3x+x^2 - 12 - 4x) = x^2-6x + 10 +x -x^2 +12 = 22-7x$$ - Example 34 Calculate \(\left(\frac{2}{3}a^2b^2-\frac{3}{2}c^4\right)^3\)
Solution$$\left(\frac{2}{3}a^2b^2-\frac{3}{2}c^4\right)^3 = \frac{8}{27}a^6b^6 - 3\cdot \left(\frac{2}{3}a^2b^2\right)^2 \cdot \frac{3}{2}c^4 + 3\cdot \frac{2}{3}a^2b^2 \cdot \left(\frac{3}{2}c^4\right)^2 - \frac{27}{8}c^{12} = $$ $$ = \frac{8}{27}a^6b^6 -2a^4b^4 c^4 + \frac{9}{2}a^2b^2c^8 - \frac{27}{8}c^{12}$$ - Example 35 Write the expression \(27x^6 - \frac{1}{8}y^3\) in the form of the product.
Solution$$ 27x^6 - \frac{1}{8}y^3 = \left(3x^2 -\frac{1}{2}y\right)\left(9x^4\right)$$ - Example 36 Write the expression \(m^2k^2 -2 m^4 k + m^6\) in the form of the product.
Solution$$m^2k^2 -2 m^4 k + m^6 = m^2\left(k^2 - 2m^2k + m^4\right) = m^2\left(k-m^2\right)^2$$ - Example 37 Write the expression \(\frac{9}{4}t^4 + 6t^2sz^3 + 4s^2z^6\) in the form of the product
Solution\begin{eqnarray}\frac{9}{4}t^4 + 6t^2sz^3 + 4s^2z^6 &=& \frac{1}{4}\left(9t^4 + 6t^2sz^3 + 4s^2z^6\right)\\ \frac{9}{4}t^4 + 24t^2sz^3 + 16s^2z^6 &=& \frac{1}{4}\left(3t^2 +4sz^3\right)\end{eqnarray}
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