Without actually calculating the cubes,find t | vedclass

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

Question

(D) We use the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.Given expression: $(31)^3 + (-16)^3 + (-15)^3$.Let $x = 31$,$y = -

Vedclass · Accepted Answer

$22320$

Vedclass · Answer

(D) We use the algebraic identity: If $x + y + z = 0$,then $x^3 + y^3 + z^3 = 3xyz$.Given expression: $(31)^3 + (-16)^3 + (-15)^3$.Let $x = 31$,$y = -16$,and $z = -15$.Check the sum: $x + y + z = 31 + (-16) + (-15) = 31 - 31 = 0$.Since the sum is $0$,the value of the expression is $3xyz$.Value $= 3 	imes (31) 	imes (-16) 	imes (-15)$.Value $= 3 	imes 31 	imes 240$.Value $= 93 	imes 240 = 22320$.

Without actually calculating the cubes,find the value of each of the following:
$(31)^{3} - (16)^{3} - (15)^{3}$

Similar Questions

For what value of $m$ is $x^{3}-2 m x^{2}+16$ divisible by $x+2$?

Is the following expression a polynomial? Justify your answer:
$\frac{(x-2)(x-4)}{x}$

For the polynomial $p(x),$ if $p(7)=0,$ then .......... is a factor of $p(x)$.

Find $p(1)$,$p(2)$,and $p(4)$ for the following polynomial: $p(t) = t^{2} - 6t + 8$.

Classify the following as a constant,linear,quadratic,or cubic polynomial: $2-x^{2}+x^{3}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Without actually calculating the cubes,find the value of each of the following:$(31)^{3} - (16)^{3} - (15)^{3}$