The value of left(frac{0.051 times 0.051 time | vedclass

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Question

(C) The given expression is of the form $\frac{a^{3} + b^{3}}{a^{2} - ab + b^{2}}$,where $a = 0.051$ and $b = 0.041$.We know the algebraic identity: $

Vedclass · Accepted Answer

$0.092$

Vedclass · Answer

(C) The given expression is of the form $\frac{a^{3} + b^{3}}{a^{2} - ab + b^{2}}$,where $a = 0.051$ and $b = 0.041$.We know the algebraic identity: $a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})$.Substituting this into the expression,we get: $\frac{(a + b)(a^{2} - ab + b^{2})}{a^{2} - ab + b^{2}} = a + b$.Now,substitute the values of $a$ and $b$: $0.051 + 0.041 = 0.092$.Therefore,the value of the expression is $0.092$.

The value of $\left(\frac{0.051 \times 0.051 \times 0.051 + 0.041 \times 0.041 \times 0.041}{0.051 \times 0.051 - 0.051 \times 0.041 + 0.041 \times 0.041}\right)$ is

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