prmo 2012 question 17



Let x1,x2,x3x_1, x_2, x_3 be the roots of the equation

x3+3x+5=0.x^3 + 3x + 5 = 0.

What is the value of the expression

(x1−1x1)(x2−1x2)(x3−1x3)\left( x_1 - \frac{1}{x_1} \right) \left( x_2 - \frac{1}{x_2} \right) \left( x_3 - \frac{1}{x_3} \right)

[PRE-RMO – 2012]

Prerequisites:
Vieta's formulas
And
(a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)


Source

Sol:
x3+0x2+3x+5=0x^3 + 0x^2 + 3x + 5 = 0 have roots x1,x2,x3x_1, x_2, x_3

⇒x1+x2+x3=0;x1x2+x2x3+x3x1=3;x1x2x3=−5\Rightarrow x_1 + x_2 + x_3 = 0 ; \quad x_1x_2 + x_2x_3 + x_3x_1 = 3 ; \quad x_1x_2x_3 = -5 (x1+1x1)(x2+1x2)(x3+1x3)⇒(x1x2+x1x2+x2x1+1x1x2)(x3+1x3)(x_1 + \frac{1}{x_1}) (x_2 + \frac{1}{x_2}) (x_3 + \frac{1}{x_3}) \Rightarrow (x_1x_2 + \frac{x_1}{x_2} + \frac{x_2}{x_1} + \frac{1}{x_1x_2})(x_3 + \frac{1}{x_3}) =(x1x2x3+x1x2x3+x3x1x2+x2x3x1+x2x1x3+x1x2x3+1x1x2x3+1x1x2x3)= (x_1x_2x_3 + \frac{x_1x_2}{x_3} + \frac{x_3x_1}{x_2} + \frac{x_2x_3}{x_1} + \frac{x_2}{x_1x_3} + \frac{x_1}{x_2x_3} + \frac{1}{x_1x_2}x_3 + \frac{1}{x_1x_2x_3}) =[(x1x2x3)2+(x1x2)2+(x1x3)2+x12+(x2x3)2+x22+x32+(1x1x2x3)2]÷x1x2x3= \left[ (x_1x_2x_3)^2 + (x_1x_2)^2 + (x_1x_3)^2 + x_1^2 + (x_2x_3)^2 + x_2^2 + x_3^2 + \left( \frac{1}{x_1x_2x_3} \right)^2 \right] \div x_1x_2x_3 =[(x1x2)2+(x2x3)2+(x1x3)2+x12+x22+x32+(1x1x2x3)2+1]÷x1x2x3(1)= \left[ (x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2 + x_1^2 + x_2^2 + x_3^2 + \left( \frac{1}{x_1x_2x_3} \right)^2 + 1 \right] \div x_1x_2x_3 \quad \text{(1)}

If
x1x2+x2x3+x3x1=3⇒(x1x2+x2x3+x3x1)2=9x_1x_2 + x_2x_3 + x_3x_1 = 3 \Rightarrow (x_1x_2 + x_2x_3 + x_3x_1)^2 = 9

⇒(x1x2)2+(x2x3)2+(x1x3)2+2x1x2x3(x1+x2+x3)=9\Rightarrow (x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2 + 2x_1x_2x_3 (x_1 + x_2 + x_3) = 9 ⇒(x1x2)2+(x2x3)2+(x1x3)2=9(11)\Rightarrow (x_1x_2)^2 + (x_2x_3)^2 + (x_1x_3)^2 = 9 \tag{11}

and

(x1+x2+x3)2=0⇒(x1+x2+x3)2=0(x_1 + x_2 + x_3)^2 = 0 \Rightarrow (x_1 + x_2 + x_3)^2 = 0 ⇒x12+x22+x32+2(x1x2+x2x3+x3x1)=0\Rightarrow x_1^2 + x_2^2 + x_3^2 + 2(x_1x_2 + x_2x_3 + x_3x_1) = 0 ⇒x12+x22+x32=−6\Rightarrow x_1^2 + x_2^2 + x_3^2 = -6

Use eq (11) in (1), we get

[9−6+(−5)2+1−5]=−295\left[ \frac{9 - 6 + (-5)^2 + 1}{-5} \right] = \frac{-29}{5}


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