Question

without actually calculating the cubes, ffind the value of (28)3+(-15)3+(-13)3​

Answer 1

456479 the correct answer of a given question

Answer 2

Answer:

16380

Explanation:

$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ is suitable.

Let $x=28$, $y=-15$, $z=-13$.

→ $x^3+y^3+z^3-3xyz=(28-15-13)(x^2+y^2+z^2-xy-yz-zx)$

→ $x^3+y^3+z^3-3xyz=0$

→ $x^3+y^3+z^3=3xyz$

Evaluating $3xyz$

$3\times 28\times (-15)\times (-13)=16380$

Hence, the value is 16380.

I didn't solve by cubing.

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