find three consecutive terms in an A.P.whose sum is -3 and the product of their cubes is512

Question

this is about algebra

Answer 1

If the AP terms are a, a+d, a+2d, their sum is 3a+3d=-3, so a+d=-1 and d=-(a+1).

The product of their cubes is a^3(a+d)^3(a+2d)^3=512.

Since a+d=-1, this equation becomes -a^3(a+2d)^3=512=2^9.

Take the cube root of each side: -a(a+2d)=2^3=8.

Substitute for d=-(a+1): -a(a-2(a+1))=8, so a(a+2)=8, and a^2+2a-8=0=(a+4)(a-2), so a=-4 or 2.

Therefore, d=-(a+1)=3 or -3.

The three terms are: -4, -1, 2 or 2, -1, -4. Their sum is -3; the product of their cubes is -64*-1*8=512.