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Factor 121+198x^6+81x^12

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$81x^{12}+198x^{6}+121$

Step-by-step explanation

Problem to solve:

$factor\left(121+198x^6+81x^{12}\right)$
1

Apply the formula: $factor\left(x\right)$, where $a^n=9$, $ab=6$, $b^n=x^{6}$ and $x=121+198x^6+81x^{12}$

$\left(9x^{6}+11\right)^{2}$
2

A binomial squared (sum) is equal to the square of the first term, plus the double product of the first by the second, plus the square of the second term. In other words: $(a+b)^2=a^2+2ab+b^2$

• Square of the first term: $\left(9x^{6}\right)^2 = \left(9x^{6}\right)^2$
• Double product of the first by the second: $2\left(9x^{6}\right)\left(11\right) = 198x^{6}$
• Square of the second term: $\left(11\right)^2 = 121$

$81x^{12}+198x^{6}+121$

$81x^{12}+198x^{6}+121$
$factor\left(121+198x^6+81x^{12}\right)$