Question

Prove that 0.75 ×0.75 ×0.75 +0.25 ×0.25 ×0.25 /0. 75 ×0.75-0. 75 ×0. 25 +0. 25 × 0. 25 = 1

Answer 1

Hi friend!!

Given,

0.75 ×0.75 ×0.75 +0.25 ×0.25 ×0.25 /0. 75 ×0.75-0. 75 ×0. 25 +0. 25 × 0. 25

Let a=0.75 b=0.25

Now, the question becomes

a³+b³/(a²-ab+b²)

we know that a³+b³=(a+b)(a²+b²-ab)

→(a+b)(a²+b²-ab)/(a²+b²-ab)

=a+b

=0.75+0.25

=1

Hence proved.

I hope this will help you ;)

Answer 2

Answer:

The given question was proved.

Step-by-step explanation:

Given, $\frac{0.75*0.75*0.75+0.25*0.25*0.25}{0.75*0.75-0.75*0.25+0.25*0.25}=1$

$\frac{0.75*0.75*0.75+0.25*0.25*0.25}{0.75*0.75-0.75*0.25+0.25*0.25}=\frac{(0.75)^3+(0.25)^3}{(0.75)^2-(0.75)(0.25)+(0.25)^2}$

                                        $=\frac{(a)^3+(b)^3}{(a)^2-(a)(b)+(b)^2}$   Where, $a=0.75, b=0.25$

                                        $=\frac{(a+b)(a^2-ab+b^2)}{(a^2-ab+b^2)}=a+b=0.75+0.25=1$

Hence proved.