Question

If a+b+c=16 and a^2+b^2+c^2=90 then find the value of a^3+b^3+c^3-3abc

Answer 1

state for typing by use of simple Question

Answer 2

Answer:

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)

(12)^2=90+2(ab+bc+ca)

144-90=2(ab+bc+ca)

54/2=ab+bc+ca

27=ab+bc+ca

a^3-b^3-c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

a^3-b^3-c^3-3abc=(a+b+c)[a^2+b^2+c^2-(ab+bc+ca)]

a^3-b^3-c^3-3abc=12X(90-(27))

a^3-b^3-c^3-3abc=12X(90-27)

a^3-b^3-c^3-3abc=12X63

a^3-b^3-c^3-3abc=756

Read more on Brainly.in - https://brainly.in/question/4043918#readmore