MCQ
The value of $\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}.$
- A$2.327$
- B$2.273$
- ✓$2$
- D$3$
✓
Answer
Correct option: C.
$2$
The given expresstion is
$\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}$
This can be written in the form
$\frac{(2.3^3)-(0.3)^3}{(2.3)^2+2.3\times0.3+(0.3)^2}$
Assume $a = 2.3$ and $b = 0.3$. then the given expression can be rewritten as $\frac{\text{a}^3-\text{b}^3}{\text{a}^2+\text{ab}+\text{b}^2}$
Recall the formula for difference of two cubes
$a^3-b^3=(a-b)\left(a^2+a b+b^2\right)$
Using the above formula, the expression becomes $\frac{\text{(a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}$
Note that both a and b are positive, unequal. so, neither $a^3-b^3$ nor any factor of it can be zero.
Therefore we can cancel the term $\left(a^2+a b+b^2\right)$ from both numerator and denominator. then the expression becomes
$\frac{(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}=\text{a}-\text{b}$
$=2.3-0.3$
$=2$
$\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}$
This can be written in the form
$\frac{(2.3^3)-(0.3)^3}{(2.3)^2+2.3\times0.3+(0.3)^2}$
Assume $a = 2.3$ and $b = 0.3$. then the given expression can be rewritten as $\frac{\text{a}^3-\text{b}^3}{\text{a}^2+\text{ab}+\text{b}^2}$
Recall the formula for difference of two cubes
$a^3-b^3=(a-b)\left(a^2+a b+b^2\right)$
Using the above formula, the expression becomes $\frac{\text{(a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}$
Note that both a and b are positive, unequal. so, neither $a^3-b^3$ nor any factor of it can be zero.
Therefore we can cancel the term $\left(a^2+a b+b^2\right)$ from both numerator and denominator. then the expression becomes
$\frac{(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}{\text{a}^2+\text{ab}+\text{b}^2}=\text{a}-\text{b}$
$=2.3-0.3$
$=2$
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