Without actually calculating the cubes,find t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $x = -12, y = 7,$ and $z = 5$.Then,$x + y + z = -12 + 7 + 5 = 0$.We know that if $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.Therefore

Vedclass · Accepted Answer

$-1260$

Vedclass · Answer

(C) Let $x = -12, y = 7,$ and $z = 5$.Then,$x + y + z = -12 + 7 + 5 = 0$.We know that if $x + y + z = 0$,then $x^{3} + y^{3} + z^{3} = 3xyz$.Therefore,$(-12)^{3} + (7)^{3} + (5)^{3} = 3[(-12)(7)(5)]$,since $(-12) + 7 + 5 = 0$.Calculating the product: $3[-420] = -1260$.Thus,$(-12)^{3} + (7)^{3} + (5)^{3} = -1260$.

Without actually calculating the cubes,find the value of each of the following: $(-12)^{3} + (7)^{3} + (5)^{3}$

Similar Questions

Factorise:
$2 \sqrt{2} a^{3} + 8 b^{3} - 27 c^{3} + 18 \sqrt{2} a b c$

Expand the following:
$\left(4-\frac{1}{3 x}\right)^{3}$

Expand $(2 x-3)(2 x+5)$.

Use suitable identities to find the products: $(3-2 x)(3+2 x)$

Expand the following:
$(3a - 5b - c)^2$

Vedclass Test Series

Exam Paper Generator

Online Exam Module