Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If a, b and c are in A.P, then the value of ${{a}^{3}}+{{c}^{3}}-8{{b}^{3}}$ is(a) $2ab$(b) $6ab$(c) $4ab$(d) None of the above

Last updated date: 17th Jun 2024
Total views: 393.3k
Views today: 10.93k
Verified
393.3k+ views
Hint: We solve this problem by using the general condition of A.P that is the sum of first and last term of A.P of three terms is equal to twice the middle term that is if $p,q,r$ are in A.P then,
$p+r=2q$
Then we apply a cube on both sides to get the required result.

We are given that the terms that are in A.P are
$a,b,c$
We know that the general condition of A.P that is the sum of first and last term of A.P of three terms is equal to twice the middle term that is if $p,q,r$ are in A.P then,
$p+r=2q$
By using the above formula to given A.P we get
$\Rightarrow a+c=2b......equation(i)$
Now, by cubing on both sides we get
$\Rightarrow {{\left( a+c \right)}^{3}}={{\left( 2b \right)}^{3}}$
We know that the formula for cube of sum of two terms that is
${{\left( x+y \right)}^{3}}={{x}^{3}}+{{y}^{3}}+3xy\left( x+y \right)$
By using the above formula we get
\begin{align} & \Rightarrow {{a}^{3}}+{{c}^{3}}+3ac\left( a+c \right)=8{{b}^{3}} \\ & \Rightarrow {{a}^{3}}+{{c}^{3}}-8{{b}^{3}}=-3ac\left( a+c \right) \\ \end{align}
By substituting the value of $\left( a+c \right)$ from equation (i) in above equation we get
\begin{align} & \Rightarrow {{a}^{3}}+{{c}^{3}}-8{{b}^{3}}=-3ac\left( 2b \right) \\ & \Rightarrow {{a}^{3}}+{{c}^{3}}-8{{b}^{3}}=-6abc \\ \end{align}
Therefore the required value is
$\therefore {{a}^{3}}+{{c}^{3}}-8{{b}^{3}}=-6abc$

So, the correct answer is “Option d”.

Note: We can solve this problem in another method that is by taking the required value.
Let us assume that the required value as
$\Rightarrow A={{a}^{3}}+{{c}^{3}}-8{{b}^{3}}$
We know that the formula of sum of cube of two terms that is
${{x}^{3}}+{{y}^{3}}={{\left( x+y \right)}^{3}}-3xy\left( x+y \right)$
By using the above formula we get
$\Rightarrow A={{\left( a+c \right)}^{3}}-3ac\left( a+c \right)-8{{b}^{3}}$
By substituting the value of $\left( a+c \right)$ from equation (i) in above equation we get
\begin{align} & \Rightarrow A={{\left( 2b \right)}^{3}}-3ac\left( 2b \right)-8{{b}^{3}} \\ & \Rightarrow A=8{{b}^{3}}-6abc+8{{b}^{3}} \\ & \Rightarrow A=-6abc \\ \end{align}
Therefore the required value is
$\therefore {{a}^{3}}+{{c}^{3}}-8{{b}^{3}}=-6abc$
So, option (d) is the correct answer.