How to solve this?

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How to solve this?

If

$a = x^{m + n} \cdot y^{l}$ $a=x^(m+n) * y^l$

$b = x^{n + l} \cdot y^{m}$ $b=x^(n+l) * y^m$

$c = x^{l + m} \cdot y^{n}$ $c=x^(l+m) * y^n$

prove that $a^{m - n} \cdot b^{n - l} \cdot c^{l - m} = 1$ $a^(m-n) * b^(n-l) * c^(l-m)=1$

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Answer 1 · 2018-04-26T13:06:37

$a^{m - n} b^{n - l} c^{l - m}$ $a^(m-n)b^(n-l)c^(l-m)$
$= {(x^{m + n} y^{l})}^{m - n} {(x^{n + l} y^{m})}^{n - l} {(x^{l + m} y^{n})}^{l - m}$ $=(x^(m+n)y^l)^(m-n)(x^(n+l)y^m)^(n-l)(x^(l+m)y^n)^(l-m)$
$= x^{(m + n) (m - n) + (n + l) (n - l) + (l + m) (l - m) \times}$ $=x^((m+n)(m-n)+(n+l)(n-l)+(l+m)(l-m) times$
$y^{l (m - n) + m (n - l) + n (l - m)}$ $y^(l(m-n)+m(n-l)+n(l-m))$
$= x^{m^{2} - n^{2} + n^{2} - l^{2} + l^{2} - m^{2}} y^{l m - ln + m n - m l + n l - n m}$ $=x^(m^2-n^2+n^2-l^2+l^2-m^2)y^(lm-ln+mn-ml+nl-nm)$
$= x^{0} y^{0} = 1$ $=x^0y^0=1$

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How to solve this?

If

$a = x^{m + n} \cdot y^{l}$ $a=x^(m+n) * y^l$

$b = x^{n + l} \cdot y^{m}$ $b=x^(n+l) * y^m$

$c = x^{l + m} \cdot y^{n}$ $c=x^(l+m) * y^n$

prove that $a^{m - n} \cdot b^{n - l} \cdot c^{l - m} = 1$ $a^(m-n) * b^(n-l) * c^(l-m)=1$

1 Answer

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If a=xm+n⋅yla=x^(m+n) * y^l b=xn+l⋅ymb=x^(n+l) * y^m c=xl+m⋅ync=x^(l+m) * y^n prove that am−n⋅bn−l⋅cl−m=1a^(m-n) * b^(n-l) * c^(l-m)=1

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Impact of this question

If

$a = x^{m + n} \cdot y^{l}$ $a=x^(m+n) * y^l$

$b = x^{n + l} \cdot y^{m}$ $b=x^(n+l) * y^m$

$c = x^{l + m} \cdot y^{n}$ $c=x^(l+m) * y^n$

prove that $a^{m - n} \cdot b^{n - l} \cdot c^{l - m} = 1$ $a^(m-n) * b^(n-l) * c^(l-m)=1$