Factor polynomial $$ 81x^{12}-1 $$
Answer
$$ 81x^{12}-1 = ( 3x^{3}-1 )( 3x^{3}+1 )( 9x^{6}+1 ) $$
Explanation
Step 1 :
Rewrite $ 81x^{12}-1 $ as:
$$ 81x^{12}-1 = (9x^{6})^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 9x^{6} $ and $ II = 1 $ , we have:
$$ 81x^{12}-1 = (9x^{6})^2 - (1)^2 = ( 9x^{6}-1 ) ( 9x^{6}+1 ) $$Step 2 :
Rewrite $ 9x^{6}-1 $ as:
$$ 9x^{6}-1 = (3x^{3})^2 - (1)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 3x^{3} $ and $ II = 1 $ , we have:
$$ 9x^{6}-1 = (3x^{3})^2 - (1)^2 = ( 3x^{3}-1 ) ( 3x^{3}+1 ) $$This page was created using
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