An algebra problem by Sabhrant Sachan

Algebra Level 4

Find the sum of all the positive integral values of x x satisfying

( x + 4 ) 3 + ( x + 5 ) 3 = ( x + 7 ) 3 + ( x − 4 ) 3 . (x+4)^3 + (x+5)^3 = (x+7)^3 + (x-4)^3 .


The answer is 5.

Sabhrant Sachan by Sabhrant Sachan , 21, India

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1 solution

Sabhrant Sachan
May 1, 2016

We have, ( x + 4 ) 3 + ( x + 5 ) 3 = ( x + 7 ) 3 + ( x − 4 ) 3 ⟹ ( x + 4 ) 3 − ( x − 4 ) 3 = ( x + 7 ) 3 − ( x + 5 ) 3 ⟹ 8 ( [ x + 4 ] 2 + [ x − 4 ] 2 + ( x + 4 ) ( x − 4 ) ) = 2 ( [ x + 7 ] 2 + [ x + 5 ] 2 + ( x + 7 ) ( x + 5 ) ) ⟹ 4 ( [ 2 x ] 2 − ( x + 4 ) ( x − 4 ) ) = ( x 2 + 49 + 14 x + x 2 + 10 x + 25 + x 2 + 12 x + 35 ) ⟹ 4 ( 3 x 2 + 16 ) = ( 3 x 2 + 36 x + 109 ) ⟹ 12 x 2 + 64 − 3 x 2 − 36 x − 109 = 0 ⟹ 9 x 2 − 36 x − 45 = 0 ⟹ x 2 − 4 x − 5 = 0 ⟹ x = − 1 , 5 ⟹ Ans = 5 \text {We have, }(x+4)^3+(x+5)^3=(x+7)^3+(x-4)^3 \\ \implies (x+4)^3-(x-4)^3=(x+7)^3-(x+5)^3 \\ \implies 8([x+4]^2+[x-4]^2+(x+4)(x-4))=2([x+7]^2+[x+5]^2+(x+7)(x+5)) \\ \implies 4([2x]^2-(x+4)(x-4))=(x^2+49+14x+x^2+10x+25+x^2+12x+35)\\ \implies 4(3x^2+16)=(3x^2+36x+109) \\ \implies 12x^2+64-3x^2-36x-109=0 \\ \implies 9x^2-36x-45=0 \\ \implies x^2-4x-5=0 \\ \implies x=-1,5 \\ \implies \color{#3D99F6}{\text{Ans = 5}}

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