Question

If the coefficient of $x^{15}$ in the expansion of $\left(a x^3+\frac{1}{b x^{1 / 3}}\right)^{15}$ is equal to the coefficient of $x^{-15}$ in the expansion of $\left(a x^{1 / 3}-\frac{1}{b x^3}\right)^{15}$, where a and $b$ are positive real numbers, then for each such ordered pair $(a, b)$ :

• A. $a=3 b$
• B. $a=b$
• C. $ab =1$
• D. $a b=3$

Answer 1

The Correct Option is C

Step 1: Coefficient of \(x^{15}\) in \((ax^3 + \frac{1}{bx^3})^{15}\)

The general term in the expansion of \((ax^3 + \frac{1}{bx^3})^{15}\) is given by:

\[ T_{r+1} = \binom{15}{r} \cdot a^{15-r} \cdot \left(\frac{1}{b}\right)^r \cdot x^{3(15-r) - 3r}. \]

Simplify the powers of \(x\):

\[ T_{r+1} = \binom{15}{r} \cdot a^{15-r} \cdot b^{-r} \cdot x^{45 - 6r}. \]

For the coefficient of \(x^{15}\), set \(45 - 6r = 15\):

\[ 45 - 15 = 6r \implies r = 9. \]

Thus, the coefficient of \(x^{15}\) is:

\[ \binom{15}{9} \cdot a^{6} \cdot b^{-9}. \]

Step 2: Coefficient of \(x^{-15}\) in \(\left(\frac{a}{x^3} - \frac{1}{bx^3}\right)^{15}\)

The general term in the expansion of \(\left(\frac{a}{x^3} - \frac{1}{bx^3}\right)^{15}\) is given by:

\[ T_{r+1} = \binom{15}{r} \cdot \left(\frac{a}{x^3}\right)^{15-r} \cdot \left(-\frac{1}{bx^3}\right)^r. \]

Simplify the powers of \(x\):

\[ T_{r+1} = \binom{15}{r} \cdot a^{15-r} \cdot b^{-r} \cdot (-1)^r \cdot x^{-3(15-r) - 3r}. \]

The exponent of \(x\) becomes:

\[ -45 + 6r. \]

For the coefficient of \(x^{-15}\), set \(-45 + 6r = -15\):

\[ 6r = 30 \implies r = 6. \]

Thus, the coefficient of \(x^{-15}\) is:

\[ \binom{15}{6} \cdot a^{9} \cdot b^{-6}. \]

Step 3: Equating the Coefficients

Equate the coefficients of \(x^{15}\) and \(x^{-15}\):

\[ \binom{15}{9} \cdot a^{6} \cdot b^{-9} = \binom{15}{6} \cdot a^{9} \cdot b^{-6}. \]

Since \(\binom{15}{9} = \binom{15}{6}\), cancel these terms:

\[ a^{6} \cdot b^{-9} = a^{9} \cdot b^{-6}. \]

Rearranging gives:

\[ \frac{a^6}{b^6} = \frac{b^9}{a^9}. \]

Cross-multiply:

\[ a^{15} \cdot b^9 = b^{15} \cdot a^9. \]

Divide both sides by \(a^9 b^9\):

\[ a^{6} = b^{6} \implies \frac{a}{b} = 1 \implies ab = 1. \]

Conclusion

The correct ordered pair satisfies \(ab = 1\).

Answer 2

Coefficient Of$x^{15}\text{in}{\left(a{x}^3+\frac{1}{b{x}^{1/3}}\right)}^{15}$
$T_{r+1}={}^{15}{C}_r{\left(a{x}^3\right)}^{15-r}{\left(\frac{1}{b{x}^{1/3}}\right)}^r$
$45-3r-\frac{r}{3}=15$
$30=\frac{10r}{3}$
$r=9$
Coefficient of $x^{15}={}^{15}{C}_9{a}^6{b}^{-9}$
Coefficient of $x^{-15}\text{in}{\left(a{x}^{1/3}-\frac{1}{b{x}^3}\right)}^{15}$
$T_{r+1}={}^{15}{C}_r{\left(ax{}^{1/3}\right)}^{15-r}{\left(-\frac{1}{b{x}^3}\right)}^r$
$5-\frac{r}{3}-3r=-15$
$\frac{10r}{3}=20$
$r=6$
Coefficient $={}^{15}{C}_6{a}^9\times b^{-6}$
$\Rightarrow \frac{a^9}{b^6}=\frac{a^6}{b^9}$
$\Rightarrow a^3{b}^3=1\Rightarrow ab=1$