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If p^2-n^2=1 prove that :
(x^m* x^n )^m-n* (x^p*x^-n)^n+p * (x^p*x^-m)^p+m = x^2



1 Answer

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Let's begin by simplifying each of the terms in the expression on the left-hand side of the equation:

(x^m * x^n)^(m-n) = x^(m-n)*(m-n)

(x^p * x^(-n))^(n+p) = x^(n+p)*(p-n)

(x^p * x^(-m))^(p+m) = x^(p+m)*(p-m)

Now, let's substitute the given equation p^2 - n^2 = 1 into these expressions:

x^(m-n)*(p^2-n^2) = x^(m-n)

x^(n+p)*(p^2-n^2) = x^(n+p)

x^(p+m)*(p^2-n^2) = x^(p+m)

Simplifying each of these expressions using the given equation, we get:

x^(m-n) = x^(m-n)

x^(n+p) = x^(n+p)

x^(p+m) = x^(p+m)

Therefore, the left-hand side of the equation simplifies to:

x^(m-n) * x^(n+p) * x^(p+m) = x^(m-n + n+p + p+m) = x^(2m) = x^2^(m-n + n+p + p+m)

Thus, we have proved that:

(x^m * x^n)^(m-n) * (x^p * x^(-n))^(n+p) * (x^p * x^(-m))^(p+m) = x^2

which is the same as the given equation.

User Leon Gaban
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