Solve for x
x=0
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9x^{2}-6x+1=9x^{2}+1
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-1\right)^{2}.
9x^{2}-6x+1-9x^{2}=1
Subtract 9x^{2} from both sides.
-6x+1=1
Combine 9x^{2} and -9x^{2} to get 0.
-6x=1-1
Subtract 1 from both sides.
-6x=0
Subtract 1 from 1 to get 0.
x=0
Product of two numbers is equal to 0 if at least one of them is 0. Since -6 is not equal to 0, x must be equal to 0.
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