Question

23
If x+y=3 and xy=2 the value of x^3-y^3 is equal to :​

Answer 1

Answer:

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Step-by-step explanation:

Given that,

x+y = 3

xy = 2

We need to find the value of y

Using equation (II)

xy=2xy=2

x=\dfrac{2}{y}x=

y

2

Put the value of x in equation (I)

\dfrac{2}{y}+y=3

y

2

+y=3

y^2-3y+2=0y

2

−3y+2=0

y^2-2y-y+2=0y

2

−2y−y+2=0

y(y-2)-1(y-2)=0y(y−2)−1(y−2)=0

(y-1)(y-2)=0(y−1)(y−2)=0

So, the value of y will be

y=2, 1y=2,1

We need to find the value of x^3-y^3x

3

−y

3

Using given equation

=x^3-y^3=x

3

−y

3

Put the value of x

=(\dfrac{2}{y})^3-y^3=(

y

2

)

3

−y

3

Put the value of y = 1

=\dfrac{8}{1}-1=

1

8

−1

=7=7

Put the value of y = 2

=\dfrac{8}{8}-2=

8

8

−2

=-1=−1

Hence, The value of x^3-y^3x

3

−y

3

is 7 and -1.