Educational

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 =

Evaluate 48³ − 30³ − 18³ Evaluate: \[ 48^3 – 30^3 – 18^3 \] Solution: Rewrite the expression: \[ 48^3 + (-30)^3 + (-18)^3 \] Since, \[ 48-30-18=0 \] Using identity: \[ a^3+b^3+c^3=3abc \quad \text{when} \quad a+b+c=0 \] Take \[ a=48,\qquad b=-30,\qquad c=-18 \] \[ 48+(-30)+(-18)=0 \] Therefore, \[ 48^3-30^3-18^3 = 3(48)(-30)(-18) \] \[ =

Evaluate 25³ − 75³ + 50³ Evaluate: \[ 25^3 – 75^3 + 50^3 \] Solution: Rewrite the expression: \[ 25^3 + 50^3 – 75^3 \] Since, \[ 25+50=75 \] Using identity: \[ a^3+b^3+c^3=3abc \quad \text{when} \quad a+b+c=0 \] Take \[ a=25,\qquad b=50,\qquad c=-75 \] \[ 25+50-75=0 \] Therefore, \[ 25^3+50^3-75^3 = 3(25)(50)(-75) \] \[ =

Find the Product Find the Product: \[ (3x – 4y + 5z) \] \[ (9x^2 + 16y^2 + 25z^2 + 12xy – 15zx + 20yz) \] Solution: Using identity: \[ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc \] Rewrite: \[ 3x-4y+5z = 3x+5z-4y \] So, \[ a=3x,\qquad b=5z,\qquad c=-4y \] \[ (3x – 4y + 5z) \] \[ (9x^2

Find the Product Find the Product: \[ (2a – 3b – 2c) \] \[ (4a^2 + 9b^2 + 4c^2 + 6ab – 6bc + 4ca) \] Solution: Using identity: \[ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc \] Rewrite: \[ 2a-3b-2c = 2a+(-3b)+(-2c) \] So, \[ a=2a,\qquad b=-3b,\qquad c=-2c \] \[ (2a – 3b – 2c) \] \[ (4a^2

Find the Product Find the Product: \[ (4x – 3y + 2z) \] \[ (16x^2 + 9y^2 + 4z^2 + 12xy + 6yz – 8zx) \] Solution: Using identity: \[ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc \] Rewrite: \[ 4x-3y+2z = 4x+2z-3y \] So, \[ a=4x,\qquad b=2z,\qquad c=-3y \] \[ (4x – 3y + 2z) \] \[ (16x^2

Evaluate Using Identity Question: If \[ y=1 \] find the value of: \[ \left(5y + \frac{15}{y}\right) \left(25y^2 – 75 + \frac{225}{y^2}\right) \] Solution: Rearranging the terms: \[ \left(5y + \frac{15}{y}\right) \left(25y^2 – 5y\cdot\frac{15}{y} + \frac{225}{y^2}\right) \] Using identity: \[ (a+b)(a^2-ab+b^2)=a^3+b^3 \] Here, \[ a=5y,\qquad b=\frac{15}{y} \] \[ = (5y)^3 + \left(\frac{15}{y}\right)^3 \] Substituting \[ y=1

Find the Product Find the Product: \[ (3x + 2y + 2z) \] \[ (9x^2 + 4y^2 + 4z^2 – 6xy – 4yz – 6zx) \] Solution: Using identity: \[ (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc \] Here, \[ a=3x,\qquad b=2y,\qquad c=2z \] \[ (3x + 2y + 2z) \] \[ (9x^2 + 4y^2 + 4z^2 – 6xy