Question

Factorize:-27a³+b³+8c³--18abc,using identity.


I whant full answer please
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Answer 1

Answer:

=(3a+b+2c)(9a2+b2+4c2−3ab−2bc−6ac)

Step-by-step explanation:

We have 27a3+b3+8c3−18abc =(3a)3+(b)3+(2c)3−3(3a)(b)(2c)

We know that x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)

Comparing both, we get

x=3a,y=b and z=2c

Then (3a+b+2c)[(3a)2+(b)2+(2c)2−(3a)(b)−(b)(2c)−(2c)(3a)]

=(3a+b+2c)(9a2+b2+4c2−3ab−2bc−6ac)

Answer 2

Step-by-step explanation:

We have 27a³+b³+8c³-18abc

Solution:

= (3a)³+(b)³+(2c)³-3(3a)(b)(2c)

we know that x³+y³+z³-3xyz = (x+y+z)(x²+y²+z²-xy-yz-zx)

Comparing both, we get

x = 3 a

y=b

z = 2c

Then,

= (3a+b+2c)[(3a)²+(b)²+(2c)²-(3a)(b)-(b)(2c)-(2c)(3a)]

= (3a+b+2c)(9a²+b²+4c²-3ab-2bc-6ac)

Hence proved using the identity

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