What is factor m^{2a}n^{2b}-1 ?

The solution to factor m^{2a}n^{2b}-1 is (n^bm^a+1)(n^bm^a-1)

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factor m^{2a}n^{2b}-1

factor

m^{2 a} n^{2 b} - 1

(n^{b} m^{a} + 1) (n^{b} m^{a} - 1)

Solution steps

m^{2 a} n^{2 b} - 1

Rewrite

m^{2 a} n^{2 b}

(m^{1 a} n^{1 b})^{2}

= (m^{1 \cdot a} n^{1 \cdot b})^{2} - 1

Rewrite

1

1^{2}

= (m^{1 a} n^{1 b})^{2} - 1^{2}

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

(m^{1 a} n^{1 b})^{2} - 1^{2} = (m^{1 a} n^{1 b} + 1) (m^{1 a} n^{1 b} - 1)

= (m^{1 a} n^{1 b} + 1) (m^{1 a} n^{1 b} - 1)

Refine

= (n^{b} m^{a} + 1) (n^{b} m^{a} - 1)

x + y - 2 (x - y)

(5 x - 2)^{3} - 8

2 m^{2} + 2 m + m n^{'} n

3^{2} - 8 x + 4

\frac{d}{d t} ((θ) (sin (θ)))

What is factor m^{2a}n^{2b}-1 ?
The solution to factor m^{2a}n^{2b}-1 is (n^bm^a+1)(n^bm^a-1)