Question

without actually calculating the cubes find the value of (a-2b)^3+(2b-3c)^3+(3c-a)^3

Answer 1

Answer:

3(a-2b)(2b-3c)(3c-a)

Step-by-step explanation:

We know that,

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

If a + b + c = 0,

⇒ a³ + b³ + c³ - 3abc = 0 × (a²+b²+c²-ab-bc-ca)

⇒ a³ + b³ + c³ - 3abc = 0

⇒ a³ + b³ + c³ = 3abc

Here, the given expression is,

$(a-2b)^3+(2b-3c)^3+(3c-a)^3$

Since, (a-2b) + (2b-3c) + (3c-a) = a - 2b + 2b - 3c + 3c - a = 0

Thus,

$(a-2b)^3+(2b-3c)^3+(3c-a)^3=3(a-2b)(2b-3c)(3c-a)$

Answer 2

Answer:

3(a-2b)(2b-3c)(3c-a)

Step-by-step explanation:

We know that,

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

If a + b + c = 0,

⇒ a³ + b³ + c³ - 3abc = 0 × (a²+b²+c²-ab-bc-ca)

⇒ a³ + b³ + c³ - 3abc = 0

⇒ a³ + b³ + c³ = 3abc

Here, the given expression is,

(a-2b)^3+(2b-3c)^3+(3c-a)^3(a−2b)

3

+(2b−3c)

3

+(3c−a)

3

Since, (a-2b) + (2b-3c) + (3c-a) = a - 2b + 2b - 3c + 3c - a = 0

Thus,

(a-2b)^3+(2b-3c)^3+(3c-a)^3=3(a-2b)(2b-3c)(3c-a)(a−2b)

3

+(2b−3c)

3

+(3c−a) 3

=3(a−2b)(2b−3c)(3c−a)