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Question
If `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m),` prove that `x^my^nz^l = x^ny^lz^m`
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Solution
Given `x = a^(m + n), y = a^(n + l)` and `z = a^(l + m)`
Putting the values of x, y and z in `x^my^nz^l,` we get
`x^my^nz^l`
= `(a^(m + n))^m(a^(n + l))^n(a^(l + m))^l`
= `(a^(m^2 + nm))(a^(n^2 + ln))(a^(l^2 + lm))`
= `a^(m^2 + n^2 + l^2 + nm + ln + lm)`
Putting the values of x, y and z in `x^ny^lz^m,` we get
`x^ny^lz^m`
= `(a^(m + n))^n(a^(n + l))^l(a^(l + m))^m`
= `(a^(mn + n^2))(a^(nl + l^2))(a^(lm + m^2))`
= `a^(mn+n^2 + nl + l^2 + lm + m^2)`
So, `x^my^nz^l = x^ny^lz^m`.
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