Question
Simplify the following products: $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
✓
Answer
In the given problem, we have to find product of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We have been given $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ On rearranging
we get, $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(\frac{1}{2}\text{a}+3\text{b}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$ By substituting $\text{x}=\frac{1}{2}\text{a},\ \text{y}=3\text{b}$
we get, $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{2}\text{a}\Big)^2-(3\text{b})^2\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{4}\text{a}^2-9\text{b}^2\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$
$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{4}\text{a}^2\Big)^2-(9\text{b}^2)^2$
$=\frac{1}{16}\text{a}^4-81\text{b}^4$
Hence the value of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ is $\frac{1}{16}\text{a}^4-81\text{b}^4.$
We have been given $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ On rearranging
we get, $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(\frac{1}{2}\text{a}+3\text{b}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$ By substituting $\text{x}=\frac{1}{2}\text{a},\ \text{y}=3\text{b}$
we get, $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{2}\text{a}\Big)^2-(3\text{b})^2\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{4}\text{a}^2-9\text{b}^2\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
We shall use the identity $\big(\text{x}-\text{y}\big)\big(\text{x}+\text{y}\big)=\text{x}^2-\text{y}^2$
$\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$
$=\Big(\frac{1}{4}\text{a}^2\Big)^2-(9\text{b}^2)^2$
$=\frac{1}{16}\text{a}^4-81\text{b}^4$
Hence the value of $\Big(\frac{1}{2}\text{a}-3\text{b}\Big)\Big(3\text{b}+\frac{1}{2}\text{a}\Big)\Big(\frac{1}{4}\text{a}^2+9\text{b}^2\Big)$ is $\frac{1}{16}\text{a}^4-81\text{b}^4.$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
→
Express in the form of $\frac{\text{p}}{\text{q}}:0.\overline{38}+1.\overline{27}.$
→In the following, using the remainder theorem, find the remainder when $f(x)$ is divided by $g(x)$ and verify the by actual division: $f(x) = 9x^3 - 3x^2 + x - 5$, $\text{g(x)}=\text{x}-\frac{2}{3}$
→Draw the graph of the following linear equations in two variables: $\frac{\text{x}-2}{3}=\text{y}-3$
→$A B C D$ is a parallelogram. $P$ is the mid-point of $A B$. $B D$ and $C P$ intersect at $Q$ such that $C Q: Q P=3: 1$. If $\operatorname{ar}(\triangle PBQ )=10\ cm^2$, find the area of parallelogram $ABCD$ .
→The perimeter of an isosceles triangle is $42\ cm$ and its base is $\Big(\frac32\Big)$ times each of the equal side. Find the length of each of the triangle, area of the triangle and the height of the triangle.
→Find the missing frequency $(p)$ for the following distribution whose mean is $7.68$
→|
x:
|
$3$
|
$5$
|
$7$
|
$9$
|
$11$
|
$13$
|
|
f:
|
$6$
|
$8$
|
$15$
|
$p$
|
$8$
|
$4$
|
In the given figure, $O$ is the centre of the circle. If $\angle\text{ACB}=50^\circ,$ find $\angle\text{OAB}.$
→Find the length of a chord which is at a distance of $4\ cm$ from the centre of a circle of radius $6\ cm.$
→Draw the graph of the following linear equations in two variables: $\frac{\text{x}}{2}-\frac{\text{y}}{3}=2$
→