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Algebra Level 3

If a + b + c = 0 a+b+c=0 , then x a 2 b − 1 c − 1 x b 2 a − 1 c − 1 x c 2 a − 1 b − 1 = ? x^{a^2 b^{-1} c^{-1}} x^{b^2 a^{-1} c^{-1}} x^{c^2 a^{-1} b^{-1}} = \ ?

x^3 x^1/2 x^(1/a^2 b^2 c^2) x^(a^2 b^2 c^2)

Keerthi Reddy by Keerthi Reddy , 45, India

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1 solution

Sabhrant Sachan
Jun 25, 2016

x a 2 b − 1 c − 1 x b 2 a − 1 c − 1 x c 2 b − 1 a − 1 ⟹ x a 2 b c + b 2 a c + c 2 a b If a + b + c = 0 , a 3 + b 3 + c 3 = 3 a b c As a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) a 2 b c + b 2 a c + c 2 a b ⟹ 1 a b c ( a 3 + b 3 + c 3 ) = 3 Answer : x 3 x^{a^{2}b^{-1}c^{-1}}x^{b^{2}a^{-1}c^{-1}}x^{c^{2}b^{-1}a^{-1}} \implies x^{\frac{a^{2}}{bc}+\frac{b^{2}}{ac}+\frac{c^{2}}{ab} } \\ \text{If } a+b+c=0 \quad , \quad a^3+b^3+c^3=3abc \\ \text{As } \hspace{2mm} a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\\ \dfrac{a^{2}}{bc}+\dfrac{b^{2}}{ac}+\dfrac{c^{2}}{ab} \implies \dfrac{1}{abc}\left( a^3+b^3+c^3 \right) = 3 \\ \text{Answer : } \boxed{ x^3 }

1 Helpful 0 Interesting 0 Brilliant 0 Confused

How did you get x a 2 b − 1 c − 1 x a 2 b − 1 c − 1 x a 2 b − 1 c − 1 x^{a^{2}b^{-1}c^{-1}}x^{a^{2}b^{-1}c^{-1}}x^{a^{2}b^{-1}c^{-1}} and a 3 + b 3 + c 3 = 3 a b c ? a^3+b^3+c^3=3abc?

Nazmus sakib - 3 years, 8 months ago

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My apologies , i have fixed the typo and explained my reasoning properly . Thanks for the correction

Sabhrant Sachan - 3 years, 8 months ago

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