Find the value of x^{3}+y^{3}-12xy+64,when x+ | vedclass

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

Question

(B) Given expression: $x^{3}+y^{3}-12xy+64$.We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.Here,let $a

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(B) Given expression: $x^{3}+y^{3}-12xy+64$.We know the algebraic identity: $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)$.Here,let $a=x$,$b=y$,and $c=4$.Then the expression becomes $x^{3}+y^{3}+4^{3}-3(x)(y)(4) = x^{3}+y^{3}+64-12xy$.Substituting $a+b+c = x+y+4$ into the identity:Since $x+y = -4$,we have $x+y+4 = -4+4 = 0$.Therefore,$x^{3}+y^{3}-12xy+64 = (0)(x^{2}+y^{2}+16-xy-4y-4x) = 0$.

Find the value of $x^{3}+y^{3}-12xy+64$,when $x+y=-4$.

Similar Questions

Factorise $144 x^{2}-289 y^{2}$.

If $p(3) = 0$ for a polynomial $p(x)$,state one factor of $p(x)$.

Find the value of the polynomial $x^{2}-7x+12$ at $x=4$.

Classify the following as a linear,quadratic,or cubic polynomial:
$5 - 3t$

Expand $\left(\frac{x}{2}+\frac{2 y}{3}-\frac{3 z}{4}\right)^{2}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module