Question

factorise x³-y³-1-3xy

Answer 1

Hey user !!

Here is your answer !!

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We have the identity

a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Here a=x, b=-y, c=-1

Now,

x^3-y^3-1-3xy = (x-y-1) (x^2+y^2+1+xy-y+x)

Hope it is satisfactory :-)

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Answer 2

Answer:

Step-by-step explanation:

x 3 – y 3 – 1 – 3xy  

x 3 – y 3 – 1 – 3xy

= (x)3 + (-y)3 + (-1)3 – 3 x (-y) (-1)

Let,  

x = a

-y = b

-1 = c

So, the given expression is,

a3 + b3 + c3 -3abc

= (a+b+c)(a2+b2+c2-ab-bc-ca)

Putting the value of a, b, c  we get,

x 3 – y 3 – 1 – 3xy

= [x + (-y) + (-1)] [x2 + (-y)2 + (-1)2 – x (-y) – (-y)(-1) – x(-1)]

= (x-y-1)(x2+y2+1+xy-y+x)