Solve the equation
-$\sqrt [ 3 ] { 3 x ^ { 2 } + 3 x - 1 } = \sqrt [ 3 ] { x }$
A) x= $\frac { 1 } { 3 }$
B) x=-1, x= $\frac { 1 } { 3 }$
C) x=-1, x=-$\frac { 1 } { 3 }$
D) No solution

Question

Solve the equation
-$\sqrt [ 3 ] { 3 x ^ { 2 } + 3 x - 1 } = \sqrt [ 3 ] { x }$
A)  x= $\frac { 1 } { 3 }$
B)  x=-1, x= $\frac { 1 } { 3 }$
C)  x=-1, x=-$\frac { 1 } { 3 }$ 
D) No solution

Quizplus · Accepted Answer

Given the equation $\sqrt [ 3 ] { 3 x ^ { 2 } + 3 x - 1 } = \sqrt [ 3 ] { x }$, both sides are cube roots, so we can equate the insides: $3x^2 + 3x - 1 = x$. Rearranging gives $3x^2 + 2x - 1 = 0$. Factoring or using the quadratic formula, we find the solutions $x = -1$ and $x = \frac{1}{3}$.

Solve the Equation - 3x2+3x−13=x3\sqrt [ 3 ] { 3 x ^ { 2 } + 3 x - 1 } = \sqrt [ 3 ] { x }33x2+3x−1​=3x​

Solve the Equation
- $\sqrt [ 3 ] { 3 x ^ { 2 } + 3 x - 1 } = \sqrt [ 3 ] { x }$