May 2026

Find the Value Using Identity Find the Value \[ x^4+\frac{1}{x^4}=119 \] Find: \[ x^3-\frac{1}{x^3} \] Solution: Using identity: \[ \left(x^2+\frac{1}{x^2}\right)^2 = x^4+\frac{1}{x^4}+2 \] \[ \left(x^2+\frac{1}{x^2}\right)^2 = 119+2 \] \[ \left(x^2+\frac{1}{x^2}\right)^2 = 121 \] \[ x^2+\frac{1}{x^2} = 11 \] Now, \[ \left(x-\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}-2 \] \[ \left(x-\frac{1}{x}\right)^2 = 11-2 \] \[ \left(x-\frac{1}{x}\right)^2 = 9 \] \[ […]

Find Values Using Identity Find the Following Values \[ x^4+\frac{1}{x^4}=194 \] Find: \[ x^3+\frac{1}{x^3},\quad x^2+\frac{1}{x^2},\quad x+\frac{1}{x} \] Solution: Using identity: \[ \left(x^2+\frac{1}{x^2}\right)^2 = x^4+\frac{1}{x^4}+2 \] \[ \left(x^2+\frac{1}{x^2}\right)^2 = 194+2 \] \[ \left(x^2+\frac{1}{x^2}\right)^2 = 196 \] \[ x^2+\frac{1}{x^2} = 14 \] Now, \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] \[ \left(x+\frac{1}{x}\right)^2 = 14+2 \] \[ \left(x+\frac{1}{x}\right)^2 = 16

Simplify Using Identity Simplify the Following \[ (2x-5y)^3-(2x+5y)^3 \] Solution: Let \[ a=2x-5y,\quad b=2x+5y \] Using identity: \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] \[ a-b = (2x-5y)-(2x+5y) = -10y \] \[ ab = (2x-5y)(2x+5y) = 4x^2-25y^2 \] \[ a+b = 4x \] \[ a^2+ab+b^2 = (a+b)^2-3ab \] \[ = (4x)^2-3(4x^2-25y^2) \] \[ = 16x^2-12x^2+75y^2 \] \[ = 4x^2+75y^2

Simplify Using Identity Simplify the Following \[ \left(x+\frac{2}{x}\right)^3 + \left(x-\frac{2}{x}\right)^3 \] Solution: Let \[ a=x+\frac{2}{x},\quad b=x-\frac{2}{x} \] Using identity: \[ a^3+b^3=(a+b)(a^2-ab+b^2) \] \[ a+b = 2x \] \[ ab = \left(x+\frac{2}{x}\right) \left(x-\frac{2}{x}\right) = x^2-\frac{4}{x^2} \] \[ a^2-ab+b^2 = (a+b)^2-3ab \] \[ = (2x)^2 -3\left(x^2-\frac{4}{x^2}\right) \] \[ = 4x^2-3x^2+\frac{12}{x^2} \] \[ = x^2+\frac{12}{x^2} \] \[ a^3+b^3

Simplify Using Identity Simplify the Following \[ \left(\frac{x}{2}+\frac{y}{3}\right)^3 – \left(\frac{x}{2}-\frac{y}{3}\right)^3 \] Solution: Let \[ a=\frac{x}{2}+\frac{y}{3},\quad b=\frac{x}{2}-\frac{y}{3} \] Using identity: \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] \[ a-b = \frac{2y}{3} \] \[ a+b = x \] \[ ab = \left(\frac{x}{2}+\frac{y}{3}\right) \left(\frac{x}{2}-\frac{y}{3}\right) = \frac{x^2}{4}-\frac{y^2}{9} \] \[ a^2+ab+b^2 = (a+b)^2-ab \] \[ = x^2-\left(\frac{x^2}{4}-\frac{y^2}{9}\right) \] \[ = \frac{3x^2}{4}+\frac{y^2}{9} \] \[ a^3-b^3

Simplify Using Identity Simplify the Following \[ (x+3)^3+(x-3)^3 \] Solution: Using identity: \[ a^3+b^3=(a+b)(a^2-ab+b^2) \] \[ = [(x+3)+(x-3)] \left[(x+3)^2-(x+3)(x-3)+(x-3)^2\right] \] \[ = (2x)\left[(x^2+6x+9)-(x^2-9)+(x^2-6x+9)\right] \] \[ = (2x)(x^2+27) \] \[ = 2x^3+54x \] Next Question / Full Exercise

Find the Value Using Identity Find the Value of \(64x^3-125z^3\), if \(4x-5z=16\) and \(xz=12\) Solution: Using identity: \[ a^3-b^3=(a-b)^3+3ab(a-b) \] Here, \[ a=4x,\quad b=5z \] \[ a-b=16 \] \[ ab=(4x)(5z)=20xz=20(12)=240 \] \[ 64x^3-125z^3 = (16)^3+3(240)(16) \] \[ = 4096+11520 \] \[ =15616 \] Next Question / Full Exercise

Find the Value Using Identity Find the Value of \(27x^3+8y^3\), if \(3x+2y=20\) and \(xy=\frac{14}{9}\) Solution: Using identity: \[ a^3+b^3=(a+b)^3-3ab(a+b) \] Here, \[ a=3x,\quad b=2y \] \[ a+b=20 \] \[ ab=(3x)(2y)=6xy =6\left(\frac{14}{9}\right) =\frac{28}{3} \] \[ 27x^3+8y^3 = (20)^3-3\left(\frac{28}{3}\right)(20) \] \[ = 8000-560 \] \[ =7440 \] Next Question / Full Exercise

Find the Value Using Identity Find the Value \[ 3x+2y=14 \] \[ xy=8 \] Find: \[ 27x^3+8y^3 \] Solution: Using identity: \[ a^3+b^3=(a+b)^3-3ab(a+b) \] Here, \[ a=3x,\quad b=2y \] \[ a+b=14 \] \[ ab=(3x)(2y)=6xy=6(8)=48 \] \[ 27x^3+8y^3 = (14)^3-3(48)(14) \] \[ = 2744-2016 \] \[ =728 \] Next Question / Full Exercise

Calculate Values Using Identity Calculate the Following Values \[ x+\frac{1}{x}=3 \] Find: \[ x^2+\frac{1}{x^2},\quad x^3+\frac{1}{x^3},\quad x^4+\frac{1}{x^4} \] Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] \[ (3)^2 = x^2+\frac{1}{x^2}+2 \] \[ 9 = x^2+\frac{1}{x^2}+2 \] \[ x^2+\frac{1}{x^2} = 7 \] Now using identity: \[ a^3+b^3=(a+b)^3-3ab(a+b) \] \[ x^3+\frac{1}{x^3} = \left(x+\frac{1}{x}\right)^3 -3\left(x\cdot\frac{1}{x}\right)\left(x+\frac{1}{x}\right) \] \[ = (3)^3-3(1)(3)