{x^(m²-n²)}^{1/(m+n} * {x^(n²-p²)}^{1/(n+p)} * {x^(p²-m²)}^{(1/(p+m)} | Indices - Simplify | SciPiPupil

Question: Simplify: $(x^{m²-n²})^{\frac{1}{m+n}} ×$$(x^{n²-l²})^{\frac{1}{n+p}} ×$$(x^{p²-m²})^{\frac{1}{p+m}}$

Solution:

Given,

$= (x^{m²-n²})^{\frac{1}{m+n}} ×(x^{n²-l²})^{\frac{1}{n+p}} ×(x^{p²-m²})^{\frac{1}{p+m}}$

$= (x^{(m+n)(m-n)})^{\frac{1}{m+n}} ×(x^{(n+p)(n-p)})^{\frac{1}{n+p}} ×(x^{(p+m)(p-m)})^{\frac{1}{p+m}}$

$= x^{m-n} × x^{n-p} × x^{p-m}$

$= x^{(m-n) +(n-p) +(p-m)}$

$= x^{m-n+n-p+p-m}$

$= x^0$

$= 1$

= Answer

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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