Question

without actually finding the cubes factorise using suitable identity (x-y)^3 + (y-z)^3 + (z-x)^3

Answer 1

Step-by-step explanation:

We know that

a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ac)

when a + b + c = 0

a³ + b³ + c³ - 3abc = 0

=> a³ + b³ + c³ = 3abc

Let a = x - y

b = y - z

c = z - x

a + b + c = x - y + y - z + z - x

=> the above equation turns out to be 3(x - y)(y - z)(z - x)

Expand it to find the answer (-3x²y + 3xy² + 3x²z - 3y²z - 3xz² + 3yz²)

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Answer 2

☑️ Given that :

${(x - y)}^{3} + {(y - z)}^{3} + {(z - x)}^{3}$

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☑️ To Find :

Factorize the given equation by using suitable identity without actually finding the cubes.

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☑️ Property Used :

As we know that, if a + b + c = 0

[ Here,

(x - y) + (y - z) + (z - x)

= x - y + y - z + z - x

= 0 ]

Then,

$\implies$$\large{\boxed{\sf{\underline{\pink{{a}^3 + {b}^3 + {c}^3 = 3abc}}}}}$

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$\huge{\{\brown{\boxed{\boxed{\mathfrak{\underline{\underline{\orange{\bigstar{.SoLutiOn.}}}}}}}}}$

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$\therefore$ By using the above property,

$\implies {(x - y)}^{3} + {(y - z)}^{3} + {(z - x)}^{3}$

$\large{\underline{\boxed{\orange{= 3 (x - y)(y - z)(z - x) }}}}$

Hence, it is the required factorization.

We can expand it futher if it is needed.

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