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Algebra Examples
Popular Problems Algebra Solve by Factoring (x-1)^(2/3)=9
(
x
−
1
)
2
3
=
9
(x-1)23=9
Move
9
9 to the left side of the equation by subtracting it from both sides.
(
x
−
1
)
2
3
−
9
=
0
(x-1)23-9=0
Rewrite
(
x
−
1
)
2
3
(x-1)23 as
(
(
x
−
1
)
1
3
)
2
((x-1)13)2.
(
(
x
−
1
)
1
3
)
2
−
9
=
0
((x-1)13)2-9=0
Rewrite
9
9 as
3
2
32.
(
(
x
−
1
)
1
3
)
2
−
3
2
=
0
((x-1)13)2-32=0
Since both terms are perfect squares, factor using the difference of squares formula,
a
2
−
b
2
=
(
a
+
b
)
(
a
−
b
)
a2-b2=(a+b)(a-b) where
a
=
(
x
−
1
)
1
3
a=(x-1)13 and
b
=
3
b=3.
(
(
x
−
1
)
1
3
+
3
)
(
(
x
−
1
)
1
3
−
3
)
=
0
((x-1)13+3)((x-1)13-3)=0
If any individual factor on the left side of the equation is equal to
0
0, the entire expression will be equal to
0
0.
(
x
−
1
)
1
3
+
3
=
0
(x-1)13+3=0
(
x
−
1
)
1
3
−
3
=
0
(x-1)13-3=0
Set the first factor equal to
0
0 and solve.
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x
=
−
26
x=-26
Set the next factor equal to
0
0 and solve.
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x
=
28
x=28
The final solution is all the values that make
(
(
x
−
1
)
1
3
+
3
)
(
(
x
−
1
)
1
3
−
3
)
=
0
((x-1)13+3)((x-1)13-3)=0 true.
x
=
−
26
,
28