What is ((m^8-y^{12}))/((m^2+y^3)) ?

The solution to ((m^8-y^{12}))/((m^2+y^3)) is (m^4+y^6)(m^2-y^3)

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simplify ((m^8-y^{12}))/((m^2+y^3))

simplify

\frac{( m ^{8} - y ^{12} )}{( m ^{2} + y ^{3} )}

(m^{4} + y^{6}) (m^{2} - y^{3})

Solution steps

\frac{m ^{8} - y ^{12}}{m ^{2} + y ^{3}}

m^{8} = (m^{4})^{2}

y^{12} = (y^{6})^{2}

= \frac{( m ^{4} ) ^{2} - ( y ^{6} ) ^{2}}{m ^{2} + y ^{3}}

Apply Difference of Two Squares Formula:

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

(m^{4})^{2} - (y^{6})^{2} = (m^{4} + y^{6}) (m^{4} - y^{6})

= \frac{( m ^{4} + y ^{6} ) ( m ^{4} - y ^{6} )}{m ^{2} + y ^{3}}

Factor

m^{4} - y^{6} : (m^{2} + y^{3}) (m^{2} - y^{3})

= \frac{( m ^{4} + y ^{6} ) ( m ^{2} + y ^{3} ) ( m ^{2} - y ^{3} )}{m ^{2} + y ^{3}}

Cancel the common factor:

m^{2} + y^{3}

= (m^{4} + y^{6}) (m^{2} - y^{3})

\frac{( ( \frac{x}{y} ) - ( \frac{y}{x} ) )}{( x ^{2} - y ^{2} )}

(2^{n})^{m}

(w^{2} z^{3})^{4}

\frac{1}{( 3 e ^{x} + e ^{- 1} )}

\frac{7 x}{4 + 70}

What is ((m^8-y^{12}))/((m^2+y^3)) ?
The solution to ((m^8-y^{12}))/((m^2+y^3)) is (m^4+y^6)(m^2-y^3)