Question

x-y=1 xy=6 find the value of x^2+y^2​

Answer 1

Answer:

x−y=−6,xy=4

(x−y)

3

=x

3

−y

3

−3x

2

y+3xy

2

=−216

=x

3

−y

3

−3yx(x−y)=−216

x

3

−y

3

=−216+3(4)(−6)

x

3

−y

3

=−288

Step-by-step explanation:

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Answer 2

Answer:

$(a-b)^{2}$ = $a^{2} +b^{2} -2ab$

$(1)^{2}$ =  $a^{2}+ b^{2}$-2×6

1+12=$a^{2} +b^{2}$

13=$a^{2}+b^{2}$

Hence, 13 is the required answer.

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