Question

if x+y+z=0 show that x³+y³+z³=3xyz
(hunt:using identity)
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Answer 1

Answer:

We know that x³+y³+z³ - 3xyz = (x+y+z) (x²+y²+z²-xy-yx-zx)

Since x+y+z = 0,

x³+y³+z³ - 3xyz = 0 (x²+y²+z²-xy-yx-zx)

x³+y³+z³ - 3xyz = 0

=> x³+y³+z³ = 3xyz

Hence proved

Answer 2

Answer:

Proved!

Step-by-step explanation:

Given:

• x + y + z = 0

To prove:

• $x^{3} + y^{3}+z^{3}= 3xyz$

Now let's do it!

So first from the given statement let's write down the value of z,

x + y + z = 0

so,

z = -x -y

taking the - as a common,

z = -(x+y) ...(1)

Now taking the LHS of what is have to prove,

LHS

$x^{3} +y^{3}+ z^{3}$

Now in there, we are substituting the value of z in (1)

so, it becomes,

$x^{3} +y^{3} + [ -(x+y)]^{3}$

Now we are going to expand $x^{3} +y^{3}$ with the formula,

$a^{3}+ b^{3} = (a+b) (a^{2} -ab + b^{2})$

so it's now,

$(x+y)(x^{2} - xy + y^{2} ) - (x+y)^{3}$

now we are splitting the (x+y)^3 part,

$(x+y)(x^{2} - xy + y^{2} ) - (x+y) (x+y)^{2}$

On taking (x+y) as common we get,

$(x+y) [(x^{2} - xy + y^{2} ) - (x+y)^{2}]$

Now expanding the (x+y)^2 part, using

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

we get,

$(x+y) [(x^{2} - xy + y^{2} ) - (x^{2} +2xy+y^{2}) ]$

On opening the brackets,

$= (x+y) [x^{2} - xy + y^{2} - x^{2} - 2xy -y^{2}) ]$

$= (x+y) [- xy -2xy ]\\= (x+y) (-3xy)\\= - (x+y) (3xy)$

now we know that -(x+y) = z which is in the (1) statement

so substituting in here,

we get,

$= (z)(3xy)\\= 3xyz$

And that our RHS!

so LHS = RHS

Hence Proved

hope this helps you

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Gautham29:)

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