Satisfactory reals

Algebra Level pending

The sum of all real numbers t t such that ( 3 t − 9 ) 3 (3^t-9)^3 + ( 9 t − 3 ) 3 (9^t-3)^3 = ( 9 t + 3 t − 12 ) 3 (9^t+3^t-12)^3 can be represented in the form a b \frac{a}{b} where a a and b b are coprime positive integers. Find a + b a+b .


The answer is 9.

Sathvik Acharya by Sathvik Acharya , 17, India

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1 solution

Kushal Bose
Mar 18, 2017

Assume 3 t − 9 = a 3^t-9=a and 9 t − 3 = b 9^t-3=b then the equation becomes

a 3 + b 3 = ( a + b ) 3 = a 3 + b 3 + 3 a b ( a + b ) ⟹ a b ( a + b ) = 0 a^3+b^3=(a+b)^3=a^3+b^3+3ab(a+b) \\ \implies ab(a+b)=0

( 3 t − 9 ) ( 9 t − 3 ) ( 3 t + 9 t − 12 ) = 0 (3^t-9)(9^t-3)(3^t+9^t-12)=0

3 t − 9 = 0 ⟹ t = 2 3^t-9=0 \implies t=2

9 t − 3 = 0 ⟹ t = 1 2 9^t-3=0 \implies t=\dfrac{1}{2}

3 t + 9 t − 12 = 0 = > 3 2 t + 3 t − 12 ⟹ ( 3 t − 1 ) ( 3 t + 4 ) = 0 ⟹ t = 1 3^t+9^t-12=0 \\ => 3^{2t} +3^t-12 \\ \implies (3^t-1)(3^t+4)=0 \implies t=1

So, sum of all rela solutions in t = 2 , 1 , 1 / 2 t=2,1,1/2

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