Question

What is the area of a rectangular park with length $(a^2+2ab+b^2)$ units and width $(a+b)$ units $?$

• A. $a^3+3a^{2}b+3a{b}^2+b^3$
• B. $a^3+3a^{3}b+3a{b}^3+b^3$
• C. $a^3+4a^{2}b+a{b}^2+b^3$
• D. None of the above

Answer 1

The correct option is A $a^3+3a^{2}b+3a{b}^2+b^3$

Given:
$\bullet$ Length of rectangular park $=(a^2+2ab+b^2)$
$\bullet$ Width of rectangular park $=(a+b)$

To Find:
Area of rectangular park

We know,
Area of a rectangle $=$ Length $\times$ Width
$=(a^2+2ab+b^2)\times (a+b)=(a+b)(a^2+2ab+b^2)$

[Applying distributive property of multiplication: $(p+q)\times r=p\times r+q\times r$]

$=a(a^2+2ab+b^2)+b(a^2+2ab+b^2)$
$=a\cdot a^2+a\cdot 2ab+a\cdot b^2+b\cdot a^2+b\cdot 2ab+b\cdot b^2$

[Applying exponent rule for multiplication: $p^m\cdot p^n=p^{m+n}$ ]

$=a^{1+2}+2a^{1+1}b+a{b}^2+a^{2}b+2a{b}^{1+1}+b^{1+2}$
$=a^3+2a^{2}b+a{b}^2+a^{2}b+2a{b}^2+b^3$

Combining like terms:

$=a^3+(2a^{2}b+a^{2}b)+(a{b}^2+2a{b}^2)+b^3$
$=a^3+3a^{2}b+3a{b}^2+b^3$

$\therefore$ Area of the rectangular park $=a^3+3a^{2}b+3a{b}^2+b^3.$

Hence, option (a.) is the correct one.

$\underset{–}{\text{Note:}}$
Let's calculate: $(a+b)\times (a+b)$
$=(a+b)(a+b)$
$=a\cdot a+a\cdot b+b\cdot a+b\cdot b$
$=a^{1+1}+ab+ab+b^{1+1}$
$=a^2+2ab+b^2$
And, $(a+b)\times (a+b)=(a+b{)}^{1+1}=(a+b{)}^2$
$\therefore (a+b{)}^2=(a^2+2ab+b^2)$
Now, $(a^2+2ab+b^2)(a+b)=(a+b{)}^2(a+b)=(a+b{)}^{2+1}=(a+b{)}^3$
And, we already get while calculating the area of the rectangular park:
$(a^2+2ab+b^2)(a+b)=a^3+3a^{2}b+3a{b}^2+b^3$
$\Rightarrow \underset{–}{(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3}$