Question
$$(x+3)^2(x^3+3x^2+3x+1) = (x^2+6x+9)(x+1)^3$$
Answer
The equation has an infinite number of solutions.
Explanation
$$ \begin{aligned} (x+3)^2(x^3+3x^2+3x+1) &= (x^2+6x+9)(x+1)^3&& \text{simplify left and right hand side} \\[1 em](x^2+6x+9)(x^3+3x^2+3x+1) &= (x^2+6x+9)(x^3+3x^2+3x+1)&& \\[1 em]x^5+9x^4+30x^3+46x^2+33x+9 &= x^5+9x^4+30x^3+46x^2+33x+9&& \text{move all terms to the left hand side } \\[1 em]x^5+9x^4+30x^3+46x^2+33x+9-x^5-9x^4-30x^3-46x^2-33x-9 &= 0&& \text{simplify left side} \\[1 em]x^5+9x^4+30x^3+46x^2+33x+9-x^5-9x^4-30x^3-46x^2-33x-9 &= 0&& \\[1 em]0 &= 0&& \\[1 em] \end{aligned} $$
Since the statement $ \color{blue}{ 0 = 0 } $ is TRUE for any value of $ x $, we conclude that the equation has infinitely many solutions.
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