Question

8xcube-bcube-12a square*b+6ab square

Answer 1

Answer:

Step-by-step explanation:

using the algebraic identity:-

x³ + y³ + 3x²y + 3xy² = ( x + y )³

Now ,

8a³ + b³ + 12a²b + 6ab²

= ( 2a )³ + b³ + 3 × ( 2a )² b + 3×(2a)b²

= ( 2a + b )³

= ( 2a + b )( 2a + b )( 2a + b )

( 2a + b ) , ( 2a + b ) , ( 2a + b ) are

factors.

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Answer 2

Answer:

The answer is - $(2x - b)^{3}$

Step-by-step explanation:

= 8$x^{3}$ - $b^{3}$ - 12$a^{2}$b + 6a$b^{2}$

To solve this follow the identity

= $(2x)^{3}$ - $(b)^{3}$ - 3$(2x)^{2}$(b) + 3(2x)$(b)^{2}$ [Using identity $(a - b)^{3}$ = $a^{3} - b^{3} - 3a^{2}b + 3ab^{2}$]

As here a = 2x and b = b (Solved above)

So,

Continue

= $(2x)^{3}$ - $(b)^{3}$ - 3$(2x)^{2}$(b) + 3(2x)$(b)^{2}$

= $(2x - b)^{3}$

Factors = (2x - b),(2x - b),(2x - b)

FOR PROVING THE ANSWER THE ANSWER RIGHT

PROOF

8$x^{3}$ - $b^{3}$ - 12$a^{2}$b + 6a$b^{2}$ = $(2x - b)^{3}$

                                    = $(2x)^{3}$ - $(b)^{3}$ - 3$(2x)^{2}$(b) + 3(2x)$(b)^{2}$  [Using identity $(a - b)^{3}$ = $a^{3} - b^{3} - 3a^{2}b + 3ab^{2}$]

8$x^{3}$ - $b^{3}$ - 12$a^{2}$b + 6a$b^{2}$ = 8

Hence proved

L.H.S = R.H.S

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