What is factor m^6n^3+p^3q^3 ?

The solution to factor m^6n^3+p^3q^3 is (m^2n+pq)(m^4n^2-m^2npq+p^2q^2)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

factor m^6n^3+p^3q^3

factor

m^{6} n^{3} + p^{3} q^{3}

(m^{2} n + pq) (m^{4} n^{2} - m^{2} n pq + p^{2} q^{2})

Solution steps

m^{6} n^{3} + p^{3} q^{3}

Rewrite

m^{6} n^{3}

(m^{2} n)^{3}

Rewrite

p^{3} q^{3}

(pq)^{3}

= (m^{2} n)^{3} + (pq)^{3}

Apply Sum of Cubes Formula:

x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2})

(m^{2} n)^{3} + (pq)^{3} = (m^{2} n + pq) ((m^{2} n)^{2} - m^{2} n pq + (pq)^{2})

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

Simplify

(m^{2} n + pq) ((m^{2} n)^{2} - m^{2} n pq + (pq)^{2}) : (m^{2} n + pq) (m^{4} n^{2} - m^{2} n pq + p^{2} q^{2})

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

3 x y - 6 x y

(\frac{π}{3}) - 3

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

13 t^{6} - 39 t^{2}

a^{5} + cos^{5} (x)

What is factor m^6n^3+p^3q^3 ?
The solution to factor m^6n^3+p^3q^3 is (m^2n+pq)(m^4n^2-m^2npq+p^2q^2)