Question

Factorize the following by using a suitable identity: (a) a³+b³-8c³+ 6abc (b) (a/b)³+ (b/c)³+ (c/a)³ -3 (c) 8x³ -27y³ + 125 z³+ 90xyz​

Answer 1

Answer:

your answer is given in the above photo

Answer 2

Step-by-step explanation:

a. We know the identity a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca)

Using the above identity taking a=a,b=−b and c=2c, the equation a3−b3+8c3+6abc can be factorised as follows:

a3−b3+8c3+6abc=(a3)+(−b)3+(2c)3−3(a)(−b)(2c)

=[a+(−b)+(2c)][a2+(−b)2+(2c)2−(a×−b)−(−b×2c)−(2c×a)]

=(a−b+2c)(a2+b2+4c2+ab+2bc−2ca)

Hence, a3−b3+

b. We know, (a−b)3=a3−b3−3a2b+3ab2

⟹  (a−b)3=a3−b3−3ab(a−b).

Then,

(a−b)3+(b−c)3+(c−a)3

=(a3−3a2b+3ab2−b3)+(b3−3b2c+3bc2−c3)+(c3−3c2a+3ca2−a3)

=−3a2b+3ab2−3b2c+3bc2−3c2a+3ca2

=3a2(c−b)+3b2(a−c)+3c

c. 8x3+27y3+125z3−90xyz

Using,

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

Here,

a=2x

b=3y

c=5z,

Therefore,

=>(2x)3+(3y)3+(5z)3−3×2x×3y×5z=(2x+3y+5z)(4x2+9y2+25z2+6xy−15yz−10xz)