If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16.

Although this solution is mentioned in the comments to the question, I think it should be put here as an answer because of its simplicity and naturalness.

Looking at the given equation $2x^3+x^2+x-1=0$ one would quickly check possible rational roots using the Rational Root Test: $$\text{To check: }\pm 1, \pm\frac 12$$

This leads to the root $a=\frac 12$. Factoring gives now

$$2x^3+x^2+x-1 = 2\left(x-\frac 12 \right)\underbrace{(x^2+x+1)}_{=\frac{x^3-1}{x-1}}$$

Hence, the other two roots are the complex conjugated 3rd roots of $1$:

$$b^3=c^3=1$$

Now, plugging in, we get

$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=-6\cdot 8\cdot 8 =-384$$

Tags:

Symmetric Polynomials

Algebra Precalculus

Roots

Solution Verification

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