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Harvard-MIT Mathematics Tournament

Algebra Difficulty 4.6 AIME Prove it United States

Problem:
Let xx be a real number such that x3+4x=8x^{3} + 4x = 8. Determine the value of x7+64x2x^{7} + 64x^{2}.

Solution

Solution:
For any integer n≥0n \geq 0, the given implies xn+3=−4xn+1+8xnx^{n+3} = -4x^{n+1} + 8x^{n}, so we can rewrite any such power of xx in terms of lower powers. Carrying out this process iteratively gives
x7=−4x5+8x4=8x4+16x3−32x2=16x3−64x2+64x=−64x2+128. \begin{aligned} x^{7} & = -4x^{5} + 8x^{4} \\ & = 8x^{4} + 16x^{3} - 32x^{2} \\ & = 16x^{3} - 64x^{2} + 64x \\ & = -64x^{2} + 128. \end{aligned}
Thus, our answer is 128128.

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