What is (i^{b+6}-up)(i^{b+6}+up) ?

The solution to (i^{b+6}-up)(i^{b+6}+up) is i^{2(b+6)}-u^2p^2

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simplify (i^{b+6}-up)(i^{b+6}+up)

simplify

(i^{b + 6} - u p) (i^{b + 6} + u p)

i^{2 (b + 6)} - u^{2} p^{2}

Solution steps

(i^{b + 6} - u p) (i^{b + 6} + u p)

Apply Difference of Two Squares

(i^{b + 6} - u p) (i^{b + 6} + u p) : i^{2 (b + 6)} - u^{2} p^{2}

= i^{2 (b + 6)} - u^{2} p^{2}

(cos (θ) + sin (θ))^{5}

(cos (θ) + sin (θ))^{3}

\frac{d ^{3}}{d x ^{3}} ((y) (5 x^{3} + 3 x^{2} - 2 x + 9))

\frac{x - 1}{x ( x + 1 )}

\frac{d ^{3}}{d x ^{3}} ((y) (6 x - 4 x^{3} + 2 - x^{4}))

What is (i^{b+6}-up)(i^{b+6}+up) ?
The solution to (i^{b+6}-up)(i^{b+6}+up) is i^{2(b+6)}-u^2p^2