In this equation $$ \sqrt[3]{x^2 + 15} = 2\sqrt[3]{x+1} $$ if I try to put in the third exponent both sides and I get an equation with the roots 7 and 1.Are these roots the same for the first equation? I get they are by doing the proof but a program says it's only 1. Can you help me?
2026-04-29 10:09:11.1777457351
Solving the equation $\sqrt[3]{x^2 + 15} = 2\sqrt[3]{x+1}$
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I am not sure that I follow what you are saying. $$ \sqrt[3]{x^2 + 15} = 2\sqrt[3]{x+1} $$ You could cube both sides and get $$ x^2 +15 = 8(x+1). $$ So $$ x^2 - 8x + 7 = 0. $$ This indeed has the solutions $1$ and $7$. You can verify (if you want to be sure) that both of these two solutions are solutions to your equation. For example $$ \sqrt[3]{7^2 + 15} = 4 = 2\sqrt[3]{7 + 1}. $$