Without actually calculating the cubes,find t | vedclass

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

Question

(B) We know the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.We n

Vedclass · Accepted Answer

$77760$

Vedclass · Answer

(B) We know the algebraic identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.If $x+y+z = 0$,then $x^{3}+y^{3}+z^{3} = 3xyz$.We need to find the value of $48^{3}-30^{3}-18^{3}$,which can be written as $48^{3}+(-30)^{3}+(-18)^{3}$.Let $x = 48$,$y = -30$,and $z = -18$.Check the sum: $x+y+z = 48 + (-30) + (-18) = 48 - 48 = 0$.Since the sum is $0$,we can use the identity $x^{3}+y^{3}+z^{3} = 3xyz$.Therefore,$48^{3}+(-30)^{3}+(-18)^{3} = 3 	imes 48 	imes (-30) 	imes (-18)$.Calculating the product: $3 	imes 48 = 144$,and $(-30) 	imes (-18) = 540$.$144 	imes 540 = 77760$.

Without actually calculating the cubes,find the value of $48^{3}-30^{3}-18^{3}$.

Similar Questions

One of the zeroes of the polynomial $2x^2 + 7x - 4$ is

Factorise the expression: $x^{2} + \frac{y^{2}}{4} + \frac{z^{2}}{16} + xy + \frac{yz}{4} + \frac{zx}{2}$

The coefficient of $x$ in the expansion of $(x+3)^{3}$ is

If $x+2$ is a factor of $x^{3}+ax^{2}-x+30$,find the value of $a$.

State whether the following statement is true or false:
The zero of the polynomial $2x + 3$ is $\frac{3}{2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module