Question

If a – b = 3, a + b + x = 2, then the value of (a – b) $\left[ {{x}^{3}}-2a{{x}^{2}}+{{a}^{2}}x-\left( a+b \right){{b}^{2}} \right]$ is
A. 84
B. 48
C. 32
D. 36

Answer 1

Hint: Suppose the given relation in the problem as f(x) and compare the equation with the remainder theorem or Euclid’s division lemma, that is given as a = bq + r, here a number ‘a’ is getting divided by ‘b’, then quotient and remainder are found as q and r respectively are found as q and r respectively. Use the property that if ‘m’ is the root of any function then (x – m) is the factor of it.

Complete step-by-step answer:
Let us represent the function $\left( a-b \right)\left[ {{x}^{3}}-2a{{x}^{2}}+{{a}^{2}}x-\left( a+b \right){{b}^{2}} \right]$ as f(x).
So, we have
$f\left( x \right)=\left( a-b \right)\left[ {{x}^{3}}-2a{{x}^{2}}+{{a}^{2}}x-\left( a+b \right){{b}^{2}} \right]$ ………. (i)
As we know the Euclid division lemma or remainder is given as
a = bq + r …………………….(ii)
Where we divide number ‘a’ by b and hence get quotient as q and remainder as r. Now, observe the relation given in the equation (i) with the equation (ii).
We can observe that we are dividing f(x) by (a – b) and getting ‘0’ as remainder and quotient as $\left( a-b \right)\left[ {{x}^{3}}-2a{{x}^{2}}+{{a}^{2}}x-\left( a+b \right){{b}^{2}} \right]$ .
Now, we know the value of ‘a – b’ from the question is ‘3’. It means that on dividing f(x) by (x – 3), remainder will become 0 as 3 is a factor of f(x) and hence 3 will be the root of f(x) as well.
Thus, the value of x should be ‘3’ for the relation.
Now, we have relations from the problem are
a + b + x = 2 ………………….(iii)
Put x = 3 to the equation (iii) to get two equation in ‘a’ and ‘b’. so, we get equation (iii) as
a + b + 3 = 2
a – b = 3 ……………… (iv)
a + b = -1 …………………(v)
Now, add equation (iv) and (v)
a – b + a + b = 3 – 1
2a = 2
a = 1 ……………………(vi)
Put a = 1 in the equation (iv) to get value of ‘b’. so, we get
1 – b = 3
b = -2 …………………..(vii)
Now, put the values of a, b from the equations (vi) and (vii) to the equation (i) and put x = 3 as well.
So, we get
$f\left( x \right)=f\left( 3 \right)=3\left[ {{3}^{3}}-2\times 1\times {{3}^{2}}+{{1}^{2}}\times 3-\left( 1-2 \right){{\left( -2 \right)}^{2}} \right]$
= 3 [27 – 18 + 3 – (-4)]
= 3(34 - 18)
$=3\times 16$
= 48
Hence, value of $\left( a-b \right)\left[ {{x}^{3}}-2a{{x}^{2}}+{{a}^{2}}x-\left( a+b \right){{b}^{2}} \right]$ is 48.
So, option (b) is the correct answer.

Note: One may try to calculate the value of the given expression by converting the given expression in three variables x, a, b to a single variable with the help of given relations in the variables. But he/she will not get the value of it as a constant, he/she will get the value of the expression in a single variable. So, one cannot solve this question with this approach. So, one needs to observe the remainder theorem concept with the given expression.
One may go wrong if he/she takes (x + 3) as a factor of the f(x) expressed in the solution. So, as 3 is the root of f(x), so x – 3 should be a factor of it as putting x – 3 to 0, will give the value of x as ‘3’.