What is factor (2mn)^2+(m^2-n^2)^2 ?

The solution to factor (2mn)^2+(m^2-n^2)^2 is (m^2+n^2)^2

factor (2mn)^2+(m^2-n^2)^2

factor

(m^{2} + n^{2})^{2}

Solution steps

(2 mn)^{2} + (m^{2} - n^{2})^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} m^{2} n^{2} + (m^{2} - n^{2})^{2}

Factor

m^{2} - n^{2} : (m + n) (m - n)

= 2^{2} m^{2} n^{2} + ((m + n) (m - n))^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} m^{2} n^{2} + (m + n)^{2} (m - n)^{2}

2^{2} = 4

= 4 m^{2} n^{2} + (m + n)^{2} (m - n)^{2}

(m + n)^{2} : m^{2} + 2 mn + n^{2}

= 4 m^{2} n^{2} + (m^{2} + 2 mn + n^{2}) (m - n)^{2}

(m - n)^{2} : m^{2} - 2 mn + n^{2}

= 4 m^{2} n^{2} + (m^{2} + 2 mn + n^{2}) (m^{2} - 2 mn + n^{2})

Expand

(m^{2} + 2 mn + n^{2}) (m^{2} - 2 mn + n^{2}) : m^{4} - 2 m^{2} n^{2} + n^{4}

= 4 m^{2} n^{2} + m^{4} - 2 m^{2} n^{2} + n^{4}

Simplify

4 m^{2} n^{2} + m^{4} - 2 m^{2} n^{2} + n^{4} : m^{4} + 2 m^{2} n^{2} + n^{4}

= m^{4} + 2 m^{2} n^{2} + n^{4}

Break the expression into groups

= (m^{4} + m^{2} n^{2}) + (m^{2} n^{2} + n^{4})

Factor out

m^{2}

from

m^{4} + m^{2} n^{2} : m^{2} (m^{2} + n^{2})

Factor out

n^{2}

from

m^{2} n^{2} + n^{4} : n^{2} (m^{2} + n^{2})

= m^{2} (m^{2} + n^{2}) + n^{2} (m^{2} + n^{2})

Factor out common term

m^{2} + n^{2}

= (m^{2} + n^{2}) (m^{2} + n^{2})

Refine

= (m^{2} + n^{2})^{2}

What is factor (2mn)^2+(m^2-n^2)^2 ?
The solution to factor (2mn)^2+(m^2-n^2)^2 is (m^2+n^2)^2