Factorise: 27x^3 + y^3 + z^3 - 9xyz | vedclass

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Question

(A) The algebraic identity is: $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.We can rewrite the given expression as:$27x^3 +

Vedclass · Accepted Answer

$(3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$

Vedclass · Answer

(A) The algebraic identity is: $x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$.We can rewrite the given expression as:$27x^3 + y^3 + z^3 - 9xyz = (3x)^3 + (y)^3 + (z)^3 - 3(3x)(y)(z)$.Comparing this with the identity,where $x$ is replaced by $3x$,$y$ by $y$,and $z$ by $z$,we get:$= (3x + y + z)((3x)^2 + y^2 + z^2 - (3x)(y) - (y)(z) - (z)(3x))$.Simplifying the terms inside the second bracket:$= (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx)$.

Factorise: $27x^3 + y^3 + z^3 - 9xyz$

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