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Question Number 49283 by Pk1167156@gmail.com last updated on 05/Dec/18

Answered by afachri last updated on 05/Dec/18

1. a^2 + b^2 + c^2 = (a + b + c)^2 − 2(ab + bc + ac) 2(ab + bc + ac) = 9 − 9 (ab + bc + ac) = 0 2. a^3 + b^3 + c^3 = (a + b + c)^3 − 3(a + b + c)(ab + bc + ac) + 3abc 24 = 3^3 − 3(3)(0) + 3abc 3abc = −3 abc = −1 and here′s the solution for question a^4 + b^4 + c^4 = (a + b + c)^4 + 4abc(a + b + c) − (ab + bc +ac)( 4(a^2 + b^2 + c^2 ) + 6(ab + bc + ac)) = 3^4 + 4(−1)(3) − (0) = 81 − 12 = 69

1.a2+b2+c2=(a+b+c)2−2(ab+bc+ac)2(ab+bc+ac)=9−9(ab+bc+ac)=02.a3+b3+c3=(a+b+c)3−3(a+b+c)(ab+bc+ac)+3abc24=33−3(3)(0)+3abc3abc=−3abc=−1andhere′sthesolutionforquestiona4+b4+c4=(a+b+c)4+4abc(a+b+c)−(ab+bc+ac)(4(a2+b2+c2)+6(ab+bc+ac))=34+4(−1)(3)−(0)=81−12=69

Commented by Pk1167156@gmail.com last updated on 05/Dec/18

Thank you very much sir

Commented by afachri last updated on 05/Dec/18

ur welcome Sir

urwelcomeSir

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