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South-Afrika, 2011

Algebra Difficulty 4.2 AIME Prove it South Africa

If x1x_1, x2x_2 and x3x_3 are the roots of x3−ax2+bx−c=0x^3 - a x^2 + b x - c = 0, prove that
(x1−x2)2+(x2−x3)2+(x3−x1)2=2a2−6b. (x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2 = 2a^2 - 6b.

Solution

Using Vieta's formula, the sum of the roots is x1+x2+x3=ax_1 + x_2 + x_3 = a, and the sum of the pairwise products is x1x2+x2x3+x3x1=bx_1 x_2 + x_2 x_3 + x_3 x_1 = b. Starting with the right hand side:
2a2−6b=2(x1+x2+x3)2−6(x1x2+x2x3+x3x1)=2(x12+x22+x32+2(x1x2+x2x3+x3x1))−6(x1x2+x2x3+x3x1)=2(x12+x22+x32)−2(x1x2+x2x3+x3x1)=(x1−x2)2+(x2−x3)2+(x3−x1)2. \begin{aligned} 2a^2 - 6b &= 2(x_1 + x_2 + x_3)^2 - 6(x_1 x_2 + x_2 x_3 + x_3 x_1) \\ &= 2(x_1^2 + x_2^2 + x_3^2 + 2(x_1 x_2 + x_2 x_3 + x_3 x_1)) - 6(x_1 x_2 + x_2 x_3 + x_3 x_1) \\ &= 2(x_1^2 + x_2^2 + x_3^2) - 2(x_1 x_2 + x_2 x_3 + x_3 x_1) \\ &= (x_1 - x_2)^2 + (x_2 - x_3)^2 + (x_3 - x_1)^2. \end{aligned}

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