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The value of (a+1b)r(a−1b)s÷(b+1a)r(b−1a)s\left(a + \frac{1}{b}\right)^r \left(a - \frac{1}{b}\right)^s \div \left(b + \frac{1}{a}\right)^r \left(b - \frac{1}{a}\right)^s is:

A$(\frac{a}{b})^{r+s}$

B$(\frac{b}{a})^{r+s}$

C$({a}{b})^{r+s}$

D$(\frac{a^r}{b^s})$

Answer:

$(\frac{a}{b})^{r+s}$

Read Explanation:

The simplified value of the given expression is (ab)r+s\left(\frac{a}{b}\right)^{r+s}.

The given expression is:
(a+1b)r(a−1b)s(b+1a)r(b−1a)s\frac{\left(a + \frac{1}{b}\right)^r \left(a - \frac{1}{b}\right)^s}{\left(b + \frac{1}{a}\right)^r \left(b - \frac{1}{a}\right)^s}

  1. Simplify the fractional terms inside each bracket:

    • (a+1b)=ab+1b\left(a + \frac{1}{b}\right) = \frac{ab + 1}{b}

    • (a−1b)=ab−1b\left(a - \frac{1}{b}\right) = \frac{ab - 1}{b}

    • (b+1a)=ab+1a\left(b + \frac{1}{a}\right) = \frac{ab + 1}{a}

    • (b−1a)=ab−1a\left(b - \frac{1}{a}\right) = \frac{ab - 1}{a}

  2. Substitute these back into the expression:
    (ab+1b)r⋅(ab−1b)s(ab+1a)r⋅(ab−1a)s\frac{\left(\frac{ab + 1}{b}\right)^r \cdot \left(\frac{ab - 1}{b}\right)^s}{\left(\frac{ab + 1}{a}\right)^r \cdot \left(\frac{ab - 1}{a}\right)^s}

  3. Separate the numerators and denominators using exponent rules:
    (ab+1)rbr⋅(ab−1)sbs(ab+1)rar⋅(ab−1)sas\frac{\frac{(ab + 1)^r}{b^r} \cdot \frac{(ab - 1)^s}{b^s}}{\frac{(ab + 1)^r}{a^r} \cdot \frac{(ab - 1)^s}{a^s}}

  4. Cancel out the common terms (ab+1)r(ab + 1)^r and (ab−1)s(ab - 1)^s from the top and bottom:
    1br⋅bs1ar⋅as=ar⋅asbr⋅bs\frac{\frac{1}{b^r \cdot b^s}}{\frac{1}{a^r \cdot a^s}} = \frac{a^r \cdot a^s}{b^r \cdot b^s}

  5. Combine using the law of indices (xm⋅xn=xm+nx^m \cdot x^n = x^{m+n}):
    ar+sbr+s=(ab)r+s\frac{a^{r+s}}{b^{r+s}} = \mathbf{\left(\frac{a}{b}\right)^{r+s}}


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