algebra #1

Algebra Level 4

The equation 2 x 3 − 6 x 2 − 12 x − 8 = 0 2x^3-6x^2-12x-8=0 has exactly one real root, which is in the form x = a + b 3 + c 3 x=a+\sqrt[3]{b}+\sqrt[3]{c} , where a a , b b and c c are positive integers.

Find a + b + c a+b+c .


The answer is 13.

Madhav Srirangan by madhav srirangan , 20, India

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1 solution

( x + 2 ) 3 = x 3 + 6 x 2 + 12 x + 8 (x+2)^3= x^3+6x^2+12x+8 − ( x + 2 ) 3 = − x 3 − 6 x 2 − 12 x − 8 -(x+2)^3= -x^3-6x^2-12x-8 3 x 3 − ( x + 2 ) 3 = 2 x 3 − 6 x 2 − 12 x − 8 3x^3 -(x+2)^3= 2x^3-6x^2-12x-8 3 x 3 = ( x + 2 ) 3 3x^3=(x+2)^3 ( x + 2 ) x = 3 1 3 \frac{(x+2)}{x}=3^ \frac{1}{3} Solving this we get x = 2 3 1 3 − 1 x= \frac{2}{3^ \frac{1}{3} -1} Rasionalize this we get x = 1 + 3 1 3 + 9 1 3 x=1+ 3^ \frac{1}{3} + 9^ \frac{1}{3} a = 1 , b = 3 , c = 9 , a + b + c = 13 a=1 ,b=3 ,c=9, a+b+c=13

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