factorise the l^2-(m-n)^2
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Answer:
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factorise एल 2- (mn 2
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How to solve your question
Your question is
2−1(−)2
l^{2}-1(m-n)^{2}l2−1(m−n)2
Simplify
1
Expand the square
2−1(−)2
l^{2}-1(m-n)^{2}l2−1(m−n)2
2−1(−)(−)
l^{2}-1(m-n)(m-n)l2−1(m−n)(m−n)
2
Distribute
2−1(−)(−)
l^{2}-1{\color{#c92786}{(m-n)(m-n)}}l2−1(m−n)(m−n)
2−1((−)−(−))
l^{2}-1({\color{#c92786}{m(m-n)-n(m-n)}})l2−1(m(m−n)−n(m−n))
3
Distribute
2−1((−)−(−))
l^{2}-1({\color{#c92786}{m(m-n)}}-n(m-n))l2−1(m(m−n)−n(m−n))
2−1(2−−(−))
l^{2}-1({\color{#c92786}{m^{2}-mn}}-n(m-n))l2−1(m2−mn−n(m−n))
4
Distribute
2−1(2−−(−))
l^{2}-1(m^{2}-mn{\color{#c92786}{-n(m-n)}})l2−1(m2−mn−n(m−n))
2−1(2−−+2)
l^{2}-1(m^{2}-mn{\color{#c92786}{-mn+n^{2}}})l2−1(m2−mn−mn+n2)
5
Combine like terms
2−1(2−−+2)
l^{2}-1(m^{2}{\color{#c92786}{-mn}}{\color{#c92786}{-mn}}+n^{2})l2−1(m2−mn−mn+n2)
2−1(2−2+2)
l^{2}-1(m^{2}{\color{#c92786}{-2mn}}+n^{2})l2−1(m2−2mn+n2)
6
Distribute
2−1(2−2+2)
l^{2}{\color{#c92786}{-1(m^{2}-2mn+n^{2})}}l2−1(m2−2mn+n2)
2−12+2−12
l^{2}{\color{#c92786}{-1m^{2}+2mn-1n^{2}}}l2−1m2+2mn−1n2
Solution
2−12+2−12
Step-by-step explanation:
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