Question

factorise 27x^3+y^3+z^3-9xyz​

Answer 1

Answer:

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

Step-by-step explanation:

$27 {x}^{3} + {y}^{3} + {z}^{3} - 9xyz \\ \\ = (3x) ^{3} + (y) ^{3} + (z) ^{3} - 3 \times 3x \times y \times z \\ \\ (Using \: identity \: a ^{3} + b ^{3} + c ^{3} - 3abc = ( a \: + b + c)(a ^{2} + b^{2} + c ^{2} - ab - bc - ca)) \\ \\ Putting \\ a = 3x \\ b = y \\ c = z \\ \\ We \: get \: ; \\ \\ (3x + y + z)( {3x}^{2} + {y}^{2} + {z}^{2} - (3x \times y) -( y \times z) - (z \times 3x) \\ \\ = (3x + y + z)(9 {x}^{2} + {y}^{2} + {z}^{2} - 3xy - yz - 3zx)$

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Answer 2

$27x3+y3+z3−9xyz</p><p></p><p></p><p>=(3x)3+y3+z3−9xyz</p><p></p><p></p><p>=(3x)3+y3+z3−3×3x×y×z</p><p></p><p></p><p>using identity</p><p></p><p></p><p>a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)</p><p></p><p></p><p>Putting a=3x,b=y,c=z</p><p></p><p></p><p>=(3x+y+z)(9x2+y2+z2−3xy−yz−3zx).</p><p></p><p>$

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