Question

27x ^ 3 + 8y ^ 3 + z ^ 3 - 18xyz ​

Answer 1

PLS MARK BRAINLIEST hope it helps

Answer 2

The first three terms are the perfect cubes of 3x, 2y, and z. The suitable identity will be $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. Here, a, b, c are 3x, 2y, and z.

And $-3\times 3x\times 2y\times z=-18xyz$ is confirmed.

Now, the answer is $(3x+2y+z)(9x^2+4y^2+z^2-6xy-2yz-3zx)$.