May 2026

Find 4x² + 4/x² Question: If \[ x-\frac{1}{x}=\frac12 \] find: \[ 4x^2+\frac{4}{x^2} \] Solution: Using identity: \[ \left(x-\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}-2 \] Substituting the given value: \[ \left(\frac12\right)^2 = x^2+\frac{1}{x^2}-2 \] \[ \frac14 = x^2+\frac{1}{x^2}-2 \] \[ x^2+\frac{1}{x^2} = 2+\frac14 \] \[ = \frac94 \] Multiplying both sides by 4: \[ 4\left(x^2+\frac{1}{x^2}\right) = 4\times\frac94 \] \[ […]

Find x⁶ + 1/x⁶ Question: If \[ x+\frac{1}{x}=3 \] find: \[ x^6+\frac{1}{x^6} \] Solution: First find \[ x^2+\frac{1}{x^2} \] 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 find \[ x^3+\frac{1}{x^3} \] Using identity: \[ \left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3} + 3\left(x+\frac{1}{x}\right) \]

Find a² + b² Question: If \[ a+b=7 \] and \[ ab=12 \] find: \[ a^2+b^2 \] Solution: Using identity: \[ (a+b)^2 = a^2+b^2+2ab \] Substituting the given values: \[ (7)^2 = a^2+b^2+2(12) \] \[ 49 = a^2+b^2+24 \] \[ a^2+b^2 = 49-24 \] \[ =25 \] Next Question / Full Exercise

Find a² + b² Question: If \[ a-b=5 \] and \[ ab=12 \] find: \[ a^2+b^2 \] Solution: Using identity: \[ (a-b)^2 = a^2+b^2-2ab \] Substituting the given values: \[ (5)^2 = a^2+b^2-2(12) \] \[ 25 = a^2+b^2-24 \] \[ a^2+b^2 = 25+24 \] \[ =49 \] Next Question / Full Exercise

Find a³ + b³ + c³ − 3abc Question: If \[ a+b+c=9 \] and \[ a^2+b^2+c^2=35 \] find: \[ a^3+b^3+c^3-3abc \] Solution: Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] First find \[ ab+bc+ca \] Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Substituting the given values: \[ 9^2 = 35+2(ab+bc+ca) \] \[ 81 = 35+2(ab+bc+ca) \]

Find x² + 1/x² Question: If \[ x+\frac{1}{x}=3 \] find: \[ x^2+\frac{1}{x^2} \] Solution: Using identity: \[ \left(x+\frac{1}{x}\right)^2 = x^2+\frac{1}{x^2}+2 \] Substituting the given value: \[ (3)^2 = x^2+\frac{1}{x^2}+2 \] \[ 9 = x^2+\frac{1}{x^2}+2 \] \[ x^2+\frac{1}{x^2} = 9-2 \] \[ =7 \] Next Question / Full Exercise

Find a³ + b³ + c³ − 3abc Question: If \[ a+b+c=9 \] and \[ a^2+b^2+c^2=26 \] find: \[ a^3+b^3+c^3-3abc \] Solution: Using identity: \[ a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] First find \[ ab+bc+ca \] Using identity: \[ (a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) \] Substituting the given values: \[ 9^2 = 26+2(ab+bc+ca) \] \[ 81 = 26+2(ab+bc+ca) \]

Find x³ + y³ + z³ − 3xyz Question: If \[ x+y+z=8 \] and \[ xy+yz+zx=20 \] find: \[ x^3+y^3+z^3-3xyz \] Solution: Using identity: \[ x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) \] First find \[ x^2+y^2+z^2 \] Using identity: \[ (x+y+z)^2 = x^2+y^2+z^2+2(xy+yz+zx) \] Substituting the given values: \[ 8^2 = x^2+y^2+z^2+2(20) \] \[ 64 = x^2+y^2+z^2+40 \]

Evaluate (0.2)³ − (0.3)³ + (0.1)³ Evaluate: \[ (0.2)^3-(0.3)^3+(0.1)^3 \] Solution: Rewrite the expression: \[ (0.2)^3+(0.1)^3+(-0.3)^3 \] Since, \[ 0.2+0.1-0.3=0 \] Using identity: \[ a^3+b^3+c^3=3abc \quad \text{when} \quad a+b+c=0 \] Take \[ a=0.2,\qquad b=0.1,\qquad c=-0.3 \] \[ 0.2+0.1-0.3=0 \] Therefore, \[ (0.2)^3-(0.3)^3+(0.1)^3 = 3(0.2)(0.1)(-0.3) \] \[ =-0.018 \] Next Question / Full Exercise

Evaluate (1/2)³ + (1/3)³ − (5/6)³ Evaluate: \[ \left(\frac12\right)^3 + \left(\frac13\right)^3 – \left(\frac56\right)^3 \] Solution: Rewrite the expression: \[ \left(\frac12\right)^3 + \left(\frac13\right)^3 + \left(-\frac56\right)^3 \] Since, \[ \frac12+\frac13-\frac56=0 \] Using identity: \[ a^3+b^3+c^3=3abc \quad \text{when} \quad a+b+c=0 \] Take \[ a=\frac12,\qquad b=\frac13,\qquad c=-\frac56 \] \[ \frac12+\frac13-\frac56=0 \] Therefore, \[ \left(\frac12\right)^3 + \left(\frac13\right)^3 – \left(\frac56\right)^3 =