Question
The value of $\frac{(2.3)^3-0.027}{(2.3)^20.69+0.09}.$
✓
Answer
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
$a^3 - b^3 = (a - b) (a^2 + ab + b^2)$
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 $(a^2 + ab + b^2)$ 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
$a^3 - b^3 = (a - b) (a^2 + ab + b^2)$
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 $(a^2 + ab + b^2)$ 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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