Find the sum of the cubes of the solutions of x^2-12x+4=0.
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\(x^2-12x+4=0\)
\(x=6\pm \sqrt{36-4}=6\pm \sqrt{2^5}\)
\(x_1=6+2^2\sqrt{2}\\ x_2=6-2^2\sqrt{2}\)
\(x_1^3+x_2^3=1584=2^4\cdot9\cdot 11\)
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