Question

factorise- 27x3 plus y3 plus z3 - 9xyz =​

Answer 1

Answer:

$\bf \large \underline{\red{Question : - }}$

• $\sf \to \: factorise \: - 27 {x}^{3} + {y}^{3} + {z}^{3} - 9xyz =$

$\bf \large \underline{\red{given : - }}$

• $\sf \to \: \: - 27 {x}^{3} + {y}^{3} + {z}^{3} - 9xyz =$

$\bf \large \underline{\red{To \: Find: - }}$

• $\text{ \sf \: Find the factory }$

$\bf \large \underline{\red{solution: - }}$

$\sf \to \: \: - 27 {x}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ \sf \to \: - 3 \times - 3 \times - 3 \times {x}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ \sf \to \: {( - 3x)}^{3} + {y}^{3} + {z}^{3} - 3 \times 3x \times y \times z$

$\sf{using \: identity \to}$

$\sf \boxed{ \sf \red{ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)}}$

$\sf \large \red {then }: - \\ \\ \sf \: \: \: \: a \: = \: 3x \\ \\ \sf \: b = y \\ \\ \sf \:c = z$

$\sf \large \red{putting \: all \: value : - }$

$\sf \:(3x + y + z )(9 {x}^{2} + {y}^{2} + {z}^{2}- 3xy - 3yz - 3zx)$

$\\\\ \sf \red{hence }\: \: 9xyz \: and - 9xyz \: both \: have \: opposite \: sign \: so \: become \: zeros$

Answer 2

27x³+y³+z³-9xyz

=> -27x³+y³+z³-9xyz

=>-3×-3×-3×x³+y³+z³-9xyz

=>(-3x)³+y³+z³ - 3×3x × y × z

Using identity:-

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

Then,

a = 3x

b = y

c = z

Putting all the value:-

(3x + y + z) (9x² + y² + z² - 3xy - 3yz - 3zx)

9xyz and -9xyz both have opposite sign so it becomes zeros.

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