Factor polynomial $$ 100y^{6}-81 $$
Answer
$$ 100y^{6}-81 = ( 10y^{3}-9 )( 10y^{3}+9 ) $$
Explanation
Rewrite $ 100y^{6}-81 $ as:
$$ 100y^{6}-81 = (10y^{3})^2 - (9)^2 $$Now we can apply the difference of squares formula.
$$ I^2 - II^2 = (I - II)(I + II) $$After putting $ I = 10y^{3} $ and $ II = 9 $ , we have:
$$ 100y^{6}-81 = (10y^{3})^2 - (9)^2 = ( 10y^{3}-9 ) ( 10y^{3}+9 ) $$This page was created using
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