(a+1/b)^m×(a-1/b)^n/(b+1/a)^m×(b-1/a)^n=(a/b)^m+n

3 min read Jun 16, 2024

Proving the Equation: (a+1/b)^m×(a-1/b)^n/(b+1/a)^m×(b-1/a)^n=(a/b)^m+n

This article will provide a step-by-step proof of the equation:

(a+1/b)^m×(a-1/b)^n/(b+1/a)^m×(b-1/a)^n=(a/b)^m+n

Step 1: Simplifying the Left Hand Side

Let's start by simplifying the left-hand side of the equation:

(a+1/b)^m×(a-1/b)^n/(b+1/a)^m×(b-1/a)^n

We can rewrite this as:

[(a+1/b)/(b+1/a)]^m × [(a-1/b)/(b-1/a)]^n

Step 2: Further Simplification

Now, let's focus on simplifying the expressions inside the brackets. For the first bracket:

(a+1/b)/(b+1/a) = (ab+1)/b / (ab+1)/a = a/b

Similarly, for the second bracket:

(a-1/b)/(b-1/a) = (ab-1)/b / (ab-1)/a = a/b

Step 3: Substituting the Simplified Expressions

Now we can substitute these simplified expressions back into the original equation:

[(a+1/b)/(b+1/a)]^m × [(a-1/b)/(b-1/a)]^n = (a/b)^m × (a/b)^n

Step 4: Applying the Rule of Exponents

We can simplify this further using the rule of exponents which states that x^m × x^n = x^(m+n). Applying this rule, we get:

(a/b)^m × (a/b)^n = (a/b)^(m+n)

Step 5: Conclusion

Therefore, we have successfully proven that:

(a+1/b)^m×(a-1/b)^n/(b+1/a)^m×(b-1/a)^n=(a/b)^m+n

This equation holds true for any non-zero values of a and b, and any integer values of m and n.