The Value of ( 2 . 3 ) 3 − 0 . 027 ( 2 . 3 ) 2 + 0 . 69 + 0 . 09 - Mathematics

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MCQ

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

Options

  • 2

  • 3

  • 2.327

  • 2.273

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Solution

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`

  Is there an error in this question or solution?
Chapter 5: Factorisation of Algebraic Expressions - Exercise 5.6 [Page 25]

APPEARS IN

RD Sharma Mathematics for Class 9
Chapter 5 Factorisation of Algebraic Expressions
Exercise 5.6 | Q 6 | Page 25

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