What is (t-3)/(t^2+9)*(t+3)/(t^2-9) ?

The solution to (t-3)/(t^2+9)*(t+3)/(t^2-9) is 1/(t^2+9)

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simplify (t-3)/(t^2+9)*(t+3)/(t^2-9)

simplify

\frac{t - 3}{t ^{2} + 9} \cdot \frac{t + 3}{t ^{2} - 9}

\frac{1}{t ^{2} + 9}

Solution steps

\frac{t - 3}{t ^{2} + 9} \cdot \frac{t + 3}{t ^{2} - 9}

Cancel

\frac{t + 3}{t ^{2} - 9} : \frac{1}{t - 3}

= \frac{t - 3}{t ^{2} + 9} \cdot \frac{1}{t - 3}

Cross-cancel common factor:

t - 3

= \frac{1}{t ^{2} + 9}

\frac{a ^{3} b ^{5}}{a ^{4} b}

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(5 + 3)^{2} + (5 - 3)^{2}

cot (θ) + tan (θ)

20 x^{10} y^{8}

What is (t-3)/(t^2+9)*(t+3)/(t^2-9) ?
The solution to (t-3)/(t^2+9)*(t+3)/(t^2-9) is 1/(t^2+9)