Question

If b is the mean proportional of a and c then \[{{\left( a-b \right)}^{3}}:{{\left( b-c \right)}^{3}}\] equal to
(a) \[{{a}^{\dfrac{3}{2}}}:{{c}^{\dfrac{3}{2}}}\]
(b) \[{{b}^{2}}:{{c}^{2}}\]
(c) \[{{a}^{2}}:{{c}^{2}}\]
(d) \[{{a}^{3}}:{{b}^{3}}\]

Answer 1

Hint: For the given question, we will write a and b in the form of ‘c’ and any constant ‘k’ then we will substitute these terms in the given ratio which we have to find and we will get the required answer.

Complete step-by-step answer:
Here, we have been given that b is the mean proportional of a and c.
\[\Rightarrow \dfrac{a}{b}=\dfrac{b}{c}\]
Now, let the ratios be equal to a constant ‘k’.
\[\Rightarrow \dfrac{a}{b}=\dfrac{b}{c}=k\]
On cross multiplying the ratios to the constant, we get,
\[\begin{align}
  & a=bk.....\left( i \right) \\
 & b=ck....\left( ii \right) \\
\end{align}\]
On substituting the value of ‘b’ in the equation (i), we get,
\[a=\left( ck \right)k=c{{k}^{2}}.....\left( iii \right)\]
We have to find the value of \[{{\left( a-b \right)}^{3}}:{{\left( b-c \right)}^{3}}\]. Now, substituting the values of a and b from equation (ii) and (iii), we get
\[\dfrac{{{\left( a-b \right)}^{3}}}{{{\left( b-c \right)}^{3}}}=\dfrac{{{\left( c{{k}^{2}}-ck \right)}^{3}}}{{{\left( ck-c \right)}^{3}}}\]
On taking ‘ck’ as common from the numerator, we get,
\[\dfrac{{{\left( a-b \right)}^{3}}}{{{\left( b-c \right)}^{3}}}=\dfrac{{{c}^{3}}{{k}^{3}}{{\left( k-1 \right)}^{3}}}{{{\left( ck-c \right)}^{3}}}\]
Again, we will take ‘c’ as common from the denominator, we get,
\[\dfrac{{{\left( a-b \right)}^{3}}}{{{\left( b-c \right)}^{3}}}=\dfrac{{{c}^{3}}{{k}^{3}}{{\left( k-1 \right)}^{3}}}{{{c}^{3}}{{\left( k-1 \right)}^{3}}}\]
We will cancel the common term from numerator and denominator, we get,
\[\dfrac{{{\left( a-b \right)}^{3}}}{{{\left( b-c \right)}^{3}}}={{k}^{3}}\]
Now, we know that from equation (i), and (ii), the values of \[k=\dfrac{a}{b}\] and \[k=\sqrt{\dfrac{a}{c}}\] respectively.
\[\dfrac{{{\left( a-b \right)}^{3}}}{{{\left( b-c \right)}^{3}}}={{\left( \dfrac{a}{b} \right)}^{3}}={{\left( \dfrac{a}{c} \right)}^{\dfrac{3}{2}}}\]
\[\Rightarrow {{\left( a-b \right)}^{3}}:{{\left( b-c \right)}^{3}}={{a}^{3}}:{{b}^{3}}={{a}^{\dfrac{3}{2}}}:{{c}^{\dfrac{3}{2}}}\]
Hence, the value of \[{{\left( a-b \right)}^{3}}:{{\left( b-c \right)}^{3}}\] are equal to \[{{a}^{3}}:{{b}^{3}}\] and \[{{a}^{\dfrac{3}{2}}}:{{c}^{\dfrac{3}{2}}}\]
Therefore, the correct options are (a) and (d).

Note: Students shouldn’t forget the second option of the given question, two options are correct for this question. Also, be careful while taking out the common terms after substituting the values of a and b to the given ratio.