Complicated Fraction?

We first see that $177^2-175^2=2(177+175)$ . $2(177+175)=2(2\cdot176)$ . Therefore, if we just look at $\frac{177^2-175^2}{176}$ , we have a quotient of $4$ . By doing the same to $\frac{173^2-171^2}{172}$ and $\frac{169^2-167^2}{168}$ , we have a product of $4\cdot4\cdot4=\boxed{64}$ as our answer.

We know $a^{2} - b^{2} = (a+b) (a-b)$ Therefore, $\frac{(177 ^2-175 ^2)(173 ^2-171 ^2)(169 ^2-167 ^2)}{176 \times 172 \times 168}.$ $\frac{(177 +175 )(177-175)(173+171)(173 -171)(169+167 )(169-167)}{176 \times 172 \times 168}.$ $=\frac{(2 \times 176 )(2)(2 \times 172)(2)(2 \times 168 )(2)}{176 \times 172 \times 168}.$ $=2^{6}$ $\boxed{=64}$

Don't worry by looking at question's title. It's not complicated.

First, separate all the terms of the numerator and solve each and finally combine them all.

First term

\­( 177^{2} \­) - \­( 175^{2} \­)

The average of the two terms 177 and 175 is 176. It can also be expressed like this.

{\­( (176 + 1)^{2} \­) - \­( (176 - 1)^{2} \­)}

By using the formula of \­( a^{2} \­) - \­( b^{2} \­)

Please note that here our a is (176 + 1) and our b is (176 - 1)

So,

( 176 + 1 + ( 176 -1)) (176 + 1 -(176 - 1))

= ( 176 + 1 + 176 - 1) (176 + 1 - 176 + 1)

=(176 + 176) ( 1+ 1)

=( 2 * 176) * 2

= 176 * 2 * 2

Second term

\­( 173^{2} \­) - \­( 171^{2} \­)

The average of the two terms 173 and 171 is 172. It can also be expressed like this.

{\­( (172 + 1)^{2} \­) - \­( (172 - 1)^{2} \­)}

By using the formula of \­( a^{2} \­) - \­( b^{2} \­)

Please note that here our a is (172 + 1) and our b is (172 - 1)

So,

( 172 + 1 + ( 172 -1)) (172 + 1 -(172 - 1))

= ( 172 + 1 + 172 - 1) (172 + 1 - 172 + 1)

=(172 + 172) ( 1+ 1)

=( 2 * 172) * 2

= 172 * 2 * 2

* Third term *

\­( 169^{2} \­) - \­( 167^{2} \­)

The average of the two terms 169 and 167 is 168. It can also be expressed like this.

{\­( (168 + 1)^{2} \­) - \­( (168 - 1)^{2} \­)}

By using the formula of \­( a^{2} \­) - \­( b^{2} \­)

Please note that here our a is (168 + 1) and our b is (168 - 1)

So,

( 168 + 1 + ( 168 -1)) (168 + 1 -(168 - 1))

= ( 168 + 1 + 168 - 1) (168 + 1 - 168 + 1)

=(168 + 168) ( 1+ 1)

=( 2 * 168) * 2

= 168 * 2 * 2

Now, combining all the terms.

(\­( 177^{2} \­) - \­( 175^{2} \­))(\­( 173^{2} \­) - \­( 171^{2} \­))(\­( 169^{2} \­) - \­( 167^{2} \­)) divided by 176 * 172 * 168

= 2 * 2 * 176 * 2 * 2 * 172 * 2 * 2* 168 divided by 176 * 172 * 168

= 2 * 2 * 2 * 2 * 2 * 2

= 64

Hence, our final answer is 64.

$\begin{array}{l} \frac{{\left( {{{177}^2} - {{175}^2}} \right)\left( {{{173}^2} - {{171}^2}} \right)\left( {{{169}^2} - {{167}^2}} \right)}}{{176 \times 172 \times 168}} = \\ \\ a = 176\quad ...\quad a = 172\quad ...\quad a = 168\\ \frac{{{{\left( {a + 1} \right)}^2} - {{\left( {a - 1} \right)}^2}}}{a} = 4\\ \\ \frac{{\left( {{{177}^2} - {{175}^2}} \right)\left( {{{173}^2} - {{171}^2}} \right)\left( {{{169}^2} - {{167}^2}} \right)}}{{176 \times 172 \times 168}} = 4 \times 4 \times 4 = 64 \end{array}$

If we factor the expression into three fractions in which the numerator is the difference of perfect squares and the denominator is the mean of the bases of those squares:

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

then ((a+b)(a-b))/((a+b)/2)

simplifies to 2*(a-b)

and since in each case a - b = 2

each of the three fraction simplifies to 2 * 2

we now have (2 * 2)(2 * 2)(2 * 2)

or 2^6 = 64

See that all the multiples of the numerator are of the form $(n+2)^2-n^2$ Which simplifies to $4n+4$ So setting $n=175,171\;and\;167$ ,we get $704,688\;and\,672$ ,So: $\frac{(177^2-175^2)(173^2-171^2)(169^2-167^2)}{176\times172\times168}=\frac{704\times688\times672}{176\times172\times168}=\frac{176\times172\times168}{44\times43\times42}=4\times4\times4=4^3=\boxed{64}$

Looking to the denominators we find that are like (a2-b2)(c2-d2)(m2-n2) we can put them as (a+b)(a-b)(c+d)(c-d)(m+n)(m-n) ,substituting given values in numerators and denominators,we get 2X2X2X2X2X2 =64 Ans K.K.GARG,India

don't blindly multiply--use (a^2-b^2)=a+b*a-b

i done by the rule