Question

please send me with process 16 and17

Answer 1

Here is your answer ....

16) Given
x²+y²= 29
xy = 2

we have to find

(i) (x+y)
using identity of $\green{(a+b)²=a²+b²+2ab}$
i.e (x+y)²= x²+y²+2xy.........(1)

Now putting the values which are given in the question
we get ,

(x+y)²= 29+2(2)
(x+y)²= 29+4= 33
x+y= √33

(ii) similarly you have to find out (x-y)

By using the identity of $\green{(a-b)²=a²+b²-2ab}$ .......(2)

(iii) By using the identity (1)

This one can be done easily

but instead of taking a and b take it as a² and b²

i.e $\green{(a²+b²)²= a⁴+ b⁴+2a²b²}$

17) ANSWER

(i) [(x-y)(x+y)](x²+y²)(x⁴+y⁴)

= By using identity
$\green{a²-b²= (a+b)(a-b)}$
we get ,

=[ (x²-y²)(x²+y²)](x⁴+y⁴)

Now again using the same identity continue the same procedure

= (x⁴-y⁴)(x⁴+y⁴)
= (x⁴)²-(y⁴)²
= x^8- y^8

(ii) Now simply using the identities (1),(2)
we can get solution to all this parts

Hope this helps you....^_^✌