Question

ii) x3y3, -17x3y3, 24x3y3, -12x3y3​

Answer 1

Answer:

Given the equation

x+y=5 and xy=6

Find out the value of (x³-y³) is

To proof

As given the equation

y =\frac{6}{x}y=

x

6

put this in the x+y=5 equation

we get

\begin{gathered}x + \frac{6}{x} =5\\ x^{2} -5x +6 =0\\ x^{2} - 2x-3x+6 =0\\x(x-2)-3(x-2)=0\end{gathered}

x+

x

6

=5

x

2

−5x+6=0

x

2

−2x−3x+6=0

x(x−2)−3(x−2)=0

thus

(x -2) (x -3) =0

x=2,x=3

When x=2

put in x+y=5

y = 3

(x³-y³) = 8 - 27

= -19

when

x=3

put in x+y=5

(x³ -y³) = 27-8

= 19

Hence proved

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