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## 1/3(x+1)^(-2/3)=0

There is one missing detail here. It's true that x = -1 is the only candidate for the solution and would have multiplicity 2, however, you must plug it into the original equation to see if it works. It turns out that it doesn't because x+1=0 is raised to the negative power and 0p is undefined when p < 0.

Indeed, the graph of y = 1/3(x + 1)-2/3 has a point-discontinuity at the point (-1,0).

Hello Diana,

The problem is

1/3(x + 1)(-2/3) = 0

Since there's a negative exponent in the denominator so, bring it to the numerator.

[(x + 1)(2/3)] / 3 = 0   (as 1/x-m = xm)

Multiply by 3 on both sides to get rid of 3 in the denominator on the left hand side. Now, it'll look like this

3 * [(x + 1)(2/3)] / 3 = 0 * 3 (cancel the 3 on the left side)

Finally, you'll get

[(x + 1)(2/3)]  = 0

Take cube on both sides of the equation

[(x + 1)(2/3)]3= (0)3

(x + 1)(2/3)*3 = 0    (as  (xm)n = x m.n)

Now it'll look like this

(x + 1)2 = 0 or (x + 1)(x + 1) = 0

x + 1 = 0 and x + 1 = 0

so x has two values

x = -1 and x = -1 (answer)

Hope this helps.