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Algebra Level 3

If P P be the product of roots of the given equation 27 x 3 + 21 x + 8 = 0 , 27x^{3}+21x+8=0, then find the value of 108 P 108P .

Note : Roots can be real as well as imaginary.


The answer is -32.

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Amir Raza by Amir Raza , 21, India

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3 solutions

Kay Xspre
Jan 26, 2016

Let all answers be p , q , r p, q, r , then ( x − p ) ( x − q ) ( x − r ) = x 3 − ( p + q + r ) x 2 + ( p q + q r + p r ) x − p q r (x-p)(x-q)(x-r) = x^3-(p+q+r)x^2+(pq+qr+pr)x-pqr Therefore p q r = − 8 27 → 108 P = 108 × − 8 27 = − 32 pqr = \frac{-8}{27}\rightarrow108P = 108\times\frac{-8}{27} = -32

4 Helpful 0 Interesting 0 Brilliant 0 Confused

Let roots of 27 x 3 + 21 x + 8 = 0 27x^3+21x+8=0 are α , β , γ \alpha ,\beta, \gamma .

P = α × β × γ = − d a = − 8 27 P=\alpha×\beta×\gamma=\frac{-d}{a}=\frac{-8}{27}

∴ 108 P = 108 × − 8 27 = − 32 \therefore 108P=108×\frac{-8}{27}=\boxed{-32}

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Harshi Singh
Feb 4, 2016

not of level 3!!!!! DAMM EASY

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