Question

The value of \[\frac{(2 . 3 )^3 - 0 . 027}{(2 . 3 )^2 + 0 . 69 + 0 . 09}\]

विकल्प

• 2

• 3

• 2.327

• 2.273

Answer 1

The given expression is

 \[\frac{(2 . 3 )^3 - 0 . 027}{(2 . 3 )^2 + 0 . 69 + 0 . 09}\]

This can be written in the form

  `((23)^3 - (0.3)^3)/((2.3)^2 + 2.3 xx 0.3 + (0.3)^2)`

Assume a =2.3and b = 0.3. Then the given expression can be rewritten as 

`(a^3 - b^3)/(a^2 + ab+ 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

`((a-b)(a^2 + ab + b^2))/(a^2 + ab + 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

`((a-b)(a^2 + ab + b^2))/(a^2 + ab + b^2) = a-b`

                                        ` = 2.3 - 0.3`

                                        ` = 2`