Question

a+b+c=18 and a^+b^+c 2 =24 find the value of a^3+b^3+c^3-3abc​

Answer 1

Answer:

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 = 18 × 24-(ab+bc+ca)

finding the value of ab+bc+ca

a+b+c =18

squaring both sides

( a+b+c) ^2 = 324

a^2+ b^2+c^2 +2ab +2bc +2ca =324

a^2+ b^2+c^2 + 2(ab +bc + ca) = 324

24 + 2(ab +bc + ca)= 324

2(ab +bc + ca)= 324-24=300

ab +bc +ca = 300/2= 150

Now putting the value

a^3+ b^3 + c^3 - 3abc = 18 × 24-(ab+bc+ca)

a^3+ b^3 + c^3 - 3abc = 18 × 24- 150

a^3+ b^3 + c^3 - 3abc = 432-150

a^3+ b^3 + c^3 - 3abc = 282