The rational root theorem won't help

Algebra Level 4

$\large x^3 + 3x^2 + 3x + 5=0$

If the real value of $x$ that satisfies the above equation is in the form $-a + \sqrt[3]{-b}$ , where $a$ and $b$ are positive integers and $b$ is cube-free, find $a + b$ .

The answer is 5.

2 solutions

Chew-Seong Cheong
Jun 27, 2016

Relevant wiki: Algebraic Manipulation - Rearranging

$\begin{aligned} x^3+3x^2+3x+5 & = 0 \\ x^3+3x^2+3x+1 & = -4 \\ (x+1)^3 & = -4 \\ x+1 & = \sqrt[3]{-4} \\ \implies x & = - 1 + \sqrt[3]{-4} \end{aligned}$

$\implies a+b = 1+4 = \boxed{5}$

Typo: $(x+1)^3 = -4$

Hung Woei Neoh - 4 years, 11 months ago

Thanks. I have changed it.

Chew-Seong Cheong - 4 years, 11 months ago

Great, same solution :)

Michael Fuller - 4 years, 11 months ago

Nice, same solution :)

Novril Razenda - 4 years, 11 months ago

But why is it level 4? I think it is overrated... ..I solved it in exactly 3 seconds (In my mind) !!!

abc xyz - 4 years, 11 months ago

Hey,where does the -4 come from?

Yau Lun - 4 years, 11 months ago

LHS - 4 = RHS - 4

Chew-Seong Cheong - 4 years, 11 months ago

Rocco Dalto
Aug 31, 2016

$x^3 + 3x^2 + 3x + 5 = 0$ $Let$ $x = u - 1 \implies$ $u^3 + 4 = 0 \implies u =$ $\sqrt[3]{-4}$ $\implies$ $x = -1 +$ $\sqrt[3]{-4}$ $therefore$ $a + b = 5$

The rational root theorem won't help

The answer is 5.

2 solutions

0 pending reports