Question 86

Find the value of $$\frac{(0.22)^3 - (0.11)^3}{0.0484 + 0.0242 + 0.0121}$$ = ?

Solution

$$\frac{(0.22)^3 - (0.11)^3}{0.0484 + 0.0242 + 0.0121}$$

This is of the form,

$$\frac{a^3-b^3}{a^2+ab+b^2}$$

We know, $$(a^3-b^3) = (a-b)(a^2+ab+b^2)$$

$$=\frac{(a-b)(a^2+ab+b^2)}{a^2+ab+b^2}$$

$$=(a-b)$$

$$=0.22-0.11 = 0.11$$

Option D is correct.

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