The value of frac{(2.3)^{3}-0.027}{(2.3)^{2}+ | vedclass

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Question

(C) The given expression is $\frac{(2.3)^{3}-0.027}{(2.3)^{2}+0.69+0.09}$.We can rewrite $0.027$ as $(0.3)^{3}$ and $0.09$ as $(0.3)^{2}$. Also,$0.69

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The given expression is $\frac{(2.3)^{3}-0.027}{(2.3)^{2}+0.69+0.09}$.We can rewrite $0.027$ as $(0.3)^{3}$ and $0.09$ as $(0.3)^{2}$. Also,$0.69 = (2.3)(0.3)$.So,the expression becomes $\frac{(2.3)^{3}-(0.3)^{3}}{(2.3)^{2}+(2.3)(0.3)+(0.3)^{2}}$.Using the algebraic identity $a^{3}-b^{3} = (a-b)(a^{2}+ab+b^{2})$,we have:$\frac{(a-b)(a^{2}+ab+b^{2})}{a^{2}+ab+b^{2}} = a-b$.Here,$a = 2.3$ and $b = 0.3$.Therefore,$a-b = 2.3 - 0.3 = 2$.

The value of $\frac{(2.3)^{3}-0.027}{(2.3)^{2}+0.69+0.09}$ is

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