Find the value of x^{3}+y^{3}+z^{3}-3xyz,when | vedclass

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

Question

(B) We are given the identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.First,we find the value of $x^{2}+y^{2}+z^{2}$ using the

Vedclass · Accepted Answer

$352$

Vedclass · Answer

(B) We are given the identity: $x^{3}+y^{3}+z^{3}-3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$.First,we find the value of $x^{2}+y^{2}+z^{2}$ using the identity $(x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2(xy+yz+zx)$.Substituting the given values: $16^{2} = x^{2}+y^{2}+z^{2} + 2(78)$.$256 = x^{2}+y^{2}+z^{2} + 156$.$x^{2}+y^{2}+z^{2} = 256 - 156 = 100$.Now,substitute these values into the original identity:$x^{3}+y^{3}+z^{3}-3xyz = (16)(100 - 78)$.$x^{3}+y^{3}+z^{3}-3xyz = 16(22) = 352$.

Find the value of $x^{3}+y^{3}+z^{3}-3xyz$,when $x+y+z=16$ and $xy+yz+zx=78$.

Similar Questions

What is the sum of the squares of the roots of the quadratic equation $x^2 - 3x + 1 = 0$?

If $\alpha$ and $\beta$ are the roots of the equation $4x^2 + 3x + 7 = 0$,then $\frac{1}{\alpha} + \frac{1}{\beta} =$

Solve the given two equations and select the correct answer from the given options.
$I.$ $x^{3} - 468 = 1729$
$II.$ $y^{2} - 1733 + 1564 = 0$

One root of the equation $3x^2 - 10x + 3 = 0$ is $\frac{1}{3}$. Find the other root.

The roots of the equation $ix^2 - 4x - 4i = 0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module