Question

Factorise : 27x^3+ y^3 + z^3 - 9xyz​

Answer 1

$\large{\underline{\boxed{\text{Solution:}}}}</p><p>$

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

Using identity

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

Putting a = 3x , b = y , c = z

$\implies {27x}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ \small { = (3x + y + z) \left[ {(3x)}^{2} + {y}^{2} + {z}^{2} - (3x)(y) - (y)(z) - (z)(3x)\right]}$

$\implies \boxed{\green{ ({3x}^{2} + {y}^{2} + {z}^{2} ) \left( {3x}^{2} + {y}^{2} + {z}^{2} - 3xy - yz - 3zx\right)}}$

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Answer 2

$\rule{300}2$

$\huge\tt{TO~FACTORISE:}$

• 27x³+ y³ + z³ - 9xyz

$\rule{300}2$

$\huge\tt{FACTORISING:}$

$\begin{lgathered}↪27 {x}^{3} + y {}^{3} + {z}^{3} - 9xyz \\ \\ ↪ (3x) {}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ ↪ {(3x)}^{3} + {y}^{3} + {z}^{3} - 3 \times 3x \times y \times z\end{lgathered}$

Putting a = 3x , b = y , c = z

$\begin{lgathered}↪ {27x}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ \small { ↪ (3x + y + z) \left[ {(3x)}^{2} + {y}^{2} + {z}^{2} - (3x)(y) - (y)(z) - (z)(3x)\right]}\end{lgathered}$

$\Large\tt{ ↪({3x}^{2} + {y}^{2} + {z}^{2} )}$

$\Large\tt{↪( {3x}^{2} + {y}^{2} + {z}^{2} - 3xy - yz - 3zx)}$

$\rule{300}2$