Question

If a +b+c= 12, a^2 +b^2 + c^2= 90. Find the value of a^3 +b^3 + c^3 - 3abc​

Answer 1

Step-by-step explanation:

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Answer 2

Heyya

It's Arjun From Shashi Public School

(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