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Question: The coefficient of \[{x^3}\] in \[{\left( {\sqrt {{x^5}} + \dfrac{3}{{\sqrt {{x^3}} }}} \right)^6}\]...

The coefficient of x3{x^3} in (x5+3x3)6{\left( {\sqrt {{x^5}} + \dfrac{3}{{\sqrt {{x^3}} }}} \right)^6} is
A.0
B.120
C.420
D.540

Explanation

Solution

Here we will rewrite the given expression and then compare it to the general form for the coefficient of the xm{x^m}. From there, we will get the desired values for the general form. Then we will substitute the obtained values to find the value of rr. Then we will put the value of rr in the general form of Tr+1{T_{r + 1}} and simplify it to get the value of the coefficient of x3{x^3}.

Complete step-by-step answer:
Given expression is (x5+3x3)6{\left( {\sqrt {{x^5}} + \dfrac{3}{{\sqrt {{x^3}} }}} \right)^6}.
We can write the above expression by removing the roots and write it in the exponent form. Therefore, we get
(x5+3x3)6=(x52+3x32)6{\left( {\sqrt {{x^5}} + \dfrac{3}{{\sqrt {{x^3}} }}} \right)^6} = {\left( {{x^{\dfrac{5}{2}}} + \dfrac{3}{{{x^{\dfrac{3}{2}}}}}} \right)^6}.
Now we will write the general form for the coefficient of xm{x^m} for expression (axp+bxq)n{\left( {a{x^p} + \dfrac{b}{{{x^q}}}} \right)^n}.
The general form is Tr+1=nCr(ax)nr(b)r{T_{r + 1}} = {}^n{C_r}{\left( {ax} \right)^{n - r}}{\left( b \right)^r} where, value of rr is r=(p×n)mp+qr = \dfrac{{\left( {p \times n} \right) - m}}{{p + q}}.
Therefore, by comparing (axp+bxq)n{\left( {a{x^p} + \dfrac{b}{{{x^q}}}} \right)^n} with the given expression, we get
p=52\q=32\n=6\m=3\a=1\b=3\begin{array}{l}p = \dfrac{5}{2}\\\q = \dfrac{3}{2}\\\n = 6\\\m = 3\\\a = 1\\\b = 3\end{array}
Now by putting all these value in the formula r=(p×n)mp+qr = \dfrac{{\left( {p \times n} \right) - m}}{{p + q}} to get its value, we get
r=(52×6)352+32r = \dfrac{{\left( {\dfrac{5}{2} \times 6} \right) - 3}}{{\dfrac{5}{2} + \dfrac{3}{2}}}
Simplifying the expression, we get
r=15382\Rightarrow r = \dfrac{{15 - 3}}{{\dfrac{8}{2}}}
Simplifying further, we get
r=124\Rightarrow r = \dfrac{{12}}{4}
Dividing 12 by 4, we get
r=3\Rightarrow r = 3
Now we will put the value of rr along with the other values in the general form of Tr+1{T_{r + 1}}. Therefore, we get
T3+1=6C3(x)63(3)3{T_{3 + 1}} = {}^6{C_3}{\left( x \right)^{6 - 3}}{\left( 3 \right)^3}
Now by solving this we will get the value of the coefficient of x3{x^3}, we get
T4=6×5×43×2×1(x)3×27\Rightarrow {T_4} = \dfrac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}}{\left( x \right)^3} \times 27
Simplifying the expression, we get
T4=20×(x)3×27\Rightarrow {T_4} = 20 \times {\left( x \right)^3} \times 27
Multiplying the terms, we get
T4=540×(x)3\Rightarrow {T_4} = 540 \times {\left( x \right)^3}
Hence the coefficient of x3{x^3} is 540.
So, option D is the correct option.

Note: Coefficient of a variable is the corresponding number or term which is written along with the variable in the expression. We should not confuse coefficient with the factors. Factors are the smallest numbers with which the given number is divisible and their multiplication will give the original number. A term in an expression can be a number or a variable or product of two or more variables or numbers.