Find the value of the following expression: \(\frac{(7.03)^{3}-0.027}{(7.03)^{2}+2.109+(0.3)^{2}}\)
6.73
The question requires calculating the value of the following mathematical expression:
$$ \frac{(7.03)^{3}-0.027}{(7.03)^{2}+2.109+(0.3)^{2}} $$
We can solve this problem by recognizing a common algebraic identity.
Let's examine the structure of the expression. We can identify parts that relate to the difference of cubes formula, which is $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$.
Let $a = 7.03$.
Now, consider the term $0.027$ in the numerator. We need to find a value $b$ such that $b^3 = 0.027$.
We can test $b = 0.3$: $$ (0.3)^3 = 0.3 \times 0.3 \times 0.3 = 0.09 \times 0.3 = 0.027 $$ So, we can set $b = 0.3$.
With $a = 7.03$ and $b = 0.3$, the numerator $(7.03)^{3}-0.027$ can be written as $a^3 - b^3$.
Now let's look at the denominator: $(7.03)^{2}+2.109+(0.3)^{2}$.
Using our identified values $a = 7.03$ and $b = 0.3$, we can check if the denominator matches the form $a^2 + ab + b^2$.
Therefore, the denominator is exactly $a^2 + ab + b^2$.
The original expression can be rewritten using $a$ and $b$ as:
$$ \frac{a^3 - b^3}{a^2 + ab + b^2} $$
Now, we substitute the factored form of the numerator, $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$: $$ \frac{(a - b)(a^2 + ab + b^2)}{a^2 + ab + b^2} $$
Since $a = 7.03$ and $b = 0.3$ are positive numbers, the term $a^2 + ab + b^2$ will not be zero. Thus, we can cancel the term $(a^2 + ab + b^2)$ from both the numerator and the denominator.
The expression simplifies to:
$$ a - b $$
Substitute the values of $a$ and $b$ back into the simplified expression $a - b$:
$$ 7.03 - 0.3 $$
Performing the subtraction:
$$ 7.03 - 0.30 = 6.73 $$
The value of the given expression is $6.73$.
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