Answer:
8
Step-by-step explanation:
given,
x+y=2
x^2+y^2=4
(x+y)^2-2(xy)=4. [(a+b)^2=a^2+b^2+2(ab)]
(2)^2-2(xy)=4
4-2(xy)=4
-2(xy)=0
xy=0
x^3+y^3=(x+y)^3-3(xy)(x+y)
[(a+b)^3=a^3+b^3+3(ab)(a+b)]
x^3+y^3=(2)^3-3(0)(2)
=8-0
=8
[tex]x^{3} +y^{3} =8\\[/tex]
Given that
x+y=2 ,
and
[tex]x^{2}+y^{2}=4\\we know that \\(x+y)^{2} =x^{2} +y^{2} +2xy\\or, 2^{2} =4+2xy\\xy= 0.\\Now,\\x^{3} +y^{3}=(x+y)(x^{2} +y^{2}-xy)\\ or, x^{3} + y^{3} = 2*(4-0)\\or, x^{3} + y^{3} =2*4=8[/tex]
Copyright © 2024 EHUB.TIPS team's - All rights reserved.
Answers & Comments
Answer:
8
Step-by-step explanation:
given,
x+y=2
x^2+y^2=4
(x+y)^2-2(xy)=4. [(a+b)^2=a^2+b^2+2(ab)]
(2)^2-2(xy)=4
4-2(xy)=4
-2(xy)=0
xy=0
x^3+y^3=(x+y)^3-3(xy)(x+y)
[(a+b)^3=a^3+b^3+3(ab)(a+b)]
x^3+y^3=(2)^3-3(0)(2)
=8-0
=8
Verified answer
Answer:
[tex]x^{3} +y^{3} =8\\[/tex]
Step-by-step explanation:
Given that
x+y=2 ,
and
[tex]x^{2}+y^{2}=4\\we know that \\(x+y)^{2} =x^{2} +y^{2} +2xy\\or, 2^{2} =4+2xy\\xy= 0.\\Now,\\x^{3} +y^{3}=(x+y)(x^{2} +y^{2}-xy)\\ or, x^{3} + y^{3} = 2*(4-0)\\or, x^{3} + y^{3} =2*4=8[/tex]