Question

(2a+1)cube +(a-1)cube​

Answer 1

Solution:-

=) ( 2a + 1)³ + ( a- 1)³

We know that,

[ ( a + b)³ = a³ + b³ + 3ab( a+b)

( a -b)³ = a³ - b³ - 3ab (a-b) ]

=) [ (2a)³ + 1 + 6a( 2a + 1) ] + [ a³ - 1 - 3a (a-1) ]

=) [ 8a³ + 1 + 12a² + 6a ] + [ a³ - 1 - 3a² + 3a ]

=) 8a³ + 1 + 12a² + 6a + a³ - 1 - 3a² + 3a

=) 9a³ + 9a² + 9a

=) 9a ( a² + a + 1 )

Hence Solved!

Identity Used:-

• ( a + b)³ = a³ + b³ + 3ab(a+b)
• ( a -b)³ = a³ - b³ - 3ab( a-b)

Other Identities:-

• ( a +b)² = a² + b² + 2ab
• ( a - b)² = a² + b² - 2ab
• ( a² + b² ) = ( a+b)(a-b)

Answer 2

Answer:  9a[a² + a + 1]

Step-by-step explanation:

Given,

(2a + 1)³ + (a - 1)³

Using (a³ + b³) = (a + b)(a² + b² -ab)

We have,

= (2a + 1 + a -1 )[(2a + 1)² + (a - 1)² - (2a + 1)(a - 1)]

= (3a)[(4a² + 1² + 2×2a×1) + (a² + 1 -2a) -(2a² -2a + a -1)]

= 3a[(4a² + 1 + 4a) + a² + 1 - 2a - 2a² + 2a - a + 1]

= 3a[4a² + 1 + 4a + 2 - a² - a]

= 3a[3a² + 3 + 3a]

= 3*3a[a² +  1  + a]

= 9a[a² + a + 1]

Note:- (a + b)² = a² + b² + 2ab

(a - b)² = a² + b² - 2ab

(a³ + b³) = (a + b)(a² + b² -ab)