Math, asked by purushothamanspurush, 2 months ago

multiply (m-n+p) ( m+n-p)​

Answers

Answered by suraj5070
90

 \huge {\boxed {\mathbb {QUESTION}}}

multiply

 (m-n+p) \times ( m+n-p)

 \huge {\boxed {\mathbb {ANSWER}}}

\implies (m-n+p) \times ( m+n-p)

 \implies m(m+n-p) - n( m+n-p)+p( m+n-p)

 \implies {m}^{2}+nm-pm- mn-{n}^{2}+pn+mp+np-{p}^{2}

 \implies {m}^{2}-{n}^{2}-{p}^{2}+\cancel {(nm-nm)} \cancel {(-pm+pm)} (+np+np)

 \implies{\boxed {\boxed {{m}^{2}-{n}^{2}-{p}^{2}+2np}}}

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 Identities

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

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

 {a}^{n} \times {a}^{m} ={a}^{n+m}

 \frac{{a}^{n}} {{a}^{m}} ={a}^{n-m}

 {{a}^{n}}^{m} ={a}^{nm}

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Answered by lacho150396
0

⟹(m−n+p)×(m+n−p)

\implies m(m+n-p) - n( m+n-p)+p( m+n-p)⟹m(m+n−p)−n(m+n−p)+p(m+n−p)

\implies {m}^{2}+nm-pm- mn-{n}^{2}+pn+mp+np-{p}^{2}⟹m

2

+nm−pm−mn−n

2

+pn+mp+np−p

2

\implies {m}^{2}-{n}^{2}-{p}^{2}+\cancel {(nm-nm)} \cancel {(-pm+pm)} (+np+np)⟹m

2

−n

2

−p

2

+

(nm−nm)

(−pm+pm)

(+np+np)

\implies{\boxed {\boxed {{m}^{2}-{n}^{2}-{p}^{2}+2np}}}⟹

m

2

−n

2

−p

2

+2np

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