Question

If x+y=5 and xy=6 then find value of x3-y3

Please give write answer

Answer 1

$x^{3}-y^{3}=\pm 19$

Tips:

• $(a-b)^{2}=(a+b)^{2}-4ab$

• $a^{3}-b^{3}=(a-b)^{3}+3ab(a-b)$

In this problem, we will use the above two algebraic identities to solve the problem.

Step-by-step explanation:

Given,

• $x+y=5$

• $xy=6$

Now, $x-y=\pm\sqrt{(x+y)^{2}-4xy}$

$\Rightarrow x-y=\pm\sqrt{5^{2}-4\times 6}$

$\Rightarrow x-y=\pm\sqrt{25-24}$

$\Rightarrow x-y=\pm\sqrt{1}$

$\Rightarrow x-y=\pm 1$

So, $x^{3}-y^{3}$

$=(x-y)^{3}+3xy(x-y)$

$=(\pm 1)^{3}+3\times 6\times (\pm 1)$

• since $x-y=1$ and $xy=6$

$=\pm 1 \pm 18$

$=\pm 19$

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Answer 2

x^3-y^3 = 19

Given

• x+y=5
• xy=6

To find

• x^3-y^3

solution

we are provided with the sum of two variables namely X and Y as 5 and their product as 6 and all asked to estimate the value of x cube minus y cube.

we know that from the Identity relating the given equations,

x^3-y^3 = (x-y)(x^2+ xy + y^2)

and

(x - y)^2 = (x+y)^2 -4xy

or, (x - y)^2 = 5^2 - 4(6)

or, (x - y)^2 = 25 - 24

or, (x - y)^2 = 1

or, x - y = 1 ( plus or minus may be considered at this section but for the sake of simplicity it is being neglected)

now,

(x+y)^2 = x^2 + y^2 +2xy

or, (5)^2 = x^2 + y^2 + 2(6)

or, 25 = x^2 + y^2 +12

or, x^2 + y^2 = 13

substituting all the values in the main equation as provided at the first,

x^3-y^3 = (x-y)(x^2+ xy + y^2)

or, x^3-y^3 = 1( 13 + 6)

or, x^3-y^3 = 19 ( the answer would contain plus or minus if it is considered at the previous section)

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