Square it for other

Algebra Level 3

I have a quadratic equation a x 2 + b x + c = 0 ax^2+bx +c =0 and whose one of the root is square of the other root. Then the value of b 3 + a 2 c + a c 2 b^3 + a^2c +ac^2 must be equal to which of the following numbers?

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3 a b c 3abc a b c abc a 2 b 2 c 3 a^2b^2c^3 2 b c a 2bca

Naren Bhandari by Naren Bhandari , 21, India

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1 solution

Let the roots be α , α 2 \alpha, \alpha ^2

We have α + α 2 = − b a \alpha + \alpha ^2 = -\dfrac ba and α × α 2 = c a \alpha \times \alpha ^2 = \dfrac ca

Now , Y = ( b 3 + a 2 c + a c 2 ) × a 3 a 3 = a 3 × ( b 3 a 3 + c a + c 2 a 2 ) Y=(b^3 + a^2c + ac^2 )\times \dfrac {a^3}{a^3} = a^3 \times (\dfrac {b^3}{a^3} + \dfrac ca + \dfrac {c^2}{a^2})

Y = a 3 ( − ( α + α 2 ) 3 + α 3 + ( α 3 ) 2 ) = a 3 ( − α 6 − α 3 − 3 α 3 ( α + α 2 ) + α 3 + α 6 ) Y = a^3 ( -(\alpha + \alpha ^2)^3 + \alpha ^3 + (\alpha ^3)^2)= a^3(-\alpha ^6 - \alpha ^3 - 3\alpha ^3(\alpha + \alpha ^2) + \alpha ^3 + \alpha ^6)

Y = a 3 ( − 3 × c a × − b a ) = 3 a b c Y = a^3( -3 \times \dfrac ca \times -\dfrac ba) = \boxed {3abc}

1 Helpful 0 Interesting 1 Brilliant 0 Confused

In line 3, there is small typo, I believe you multiplied by a 3 a 3 \frac{a^3}{a^3} and not ( a 3 ) ( a 3 ) (a^3)(a^3)

Sathvik Acharya - 2 years, 5 months ago

Thanks I have corrected that:)

A Former Brilliant Member - 2 years, 5 months ago

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