Answer:
It is always good if you solve first by your own.
x3=xx3=x
Subtract x2x2 from both sides to get:
x3−x2=−x2+xx3−x2=−x2+x
x(x2−x)=−1(x2−x)x(x2−x)=−1(x2−x)
Now, we know that x2x2 is not equal to xx, which means x2−xx2−x is not equal to zero. So, it’s safe to divide both sides by x2−xx2−x :
x=−1
So, the answer is -1.
Here, it was a key that since x and x square are not equal, x square - x can be cut from both sides of the equation. This results in x as minus 1 as the answer.
It is always good if you solve first by your own.
x3=xx3=x
Subtract x2x2 from both sides to get:
x3−x2=−x2+xx3−x2=−x2+x
x(x2−x)=−1(x2−x)x(x2−x)=−1(x2−x)
Now, we know that x2x2 is not equal to xx, which means x2−xx2−x is not equal to zero. So, it’s safe to divide both sides by x2−xx2−x :
x=−1
So, the answer is -1.
Here, it was a key that since x and x square are not equal, x square - x can be cut from both sides of the equation. This results in x as minus 1 as the answer.
