Question

2 * root(x + 1, 3) + 3 * root(x, 3) = 2​

Answer 1

Explicație pas cu pas:

$(a-b)^3=a^3-3a^2b+3ab^2-b^3$

$2\sqrt[3]{x+1} +3\sqrt[3]{x} =2\\\sqrt[3]{x+1} =t\\x+1=t^3 \implies x= t^3-1\\2t+3\sqrt[3]{t^3-1}}=2\\ 3\sqrt[3]{t^3-1}}=2-2t\\27(t^3-1)=8-24t+24t^3-8t^3\\-35t^3+24t^2-24t+35=0 \ \\Observam \ t=1 \ radacina \ si \ descompunem \ prin \ schema \ lui \ Horner.\\(1-t)\underbrace{(35t^2+11t+35)}_{\Delta < 0}=0$

$1-t=0\implies t=1\\\sqrt[3]{x+1} =1\\x+1=1 \implies \boxed{x=0}$