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# Cubic Identities

We shall prove it now.

Cubic Identities

I. (a + b)3 = a3 + 3a2b + 3ab2 + b3

We shall prove it now,

LHS = (a + b)3

= [(a + b)(a + b)](a + b) (expanded form)

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

= (a 2 + 2ab + b2 )(a + b) (using identity)

= a(a2 + 2ab + b2 ) + b(a 2 + 2ab + b2 ) (using distributive law)

= a 3 + 2a2b + ab2 + ba2 + 2ab2 + b3

= a3 + (2a2b + ba2 ) + (ab2 + 2ab2 ) + b3 (grouping ‘like’ terms)

= a 3 + 3a2b + 3ab 2 + b3

= RHS

(a + b)3 = a3 + 3a2b + 3ab2 + b3

Hence, we proved the cubic identity by direct multiplication.

Activity

You can visualize the geometrical proof of (a+b)3 with the help of your teacher.

II. (ab)3 = a3 – 3a2b + 3ab2b3

We can prove this identity by direct multiplication

We have (a b)3 = (a b)(a b)(a b)

= (ab)2 × (ab)

= (a 22ab + b2 )(ab)

= a(a22ab + b 2 )b(a 22ab + b2 )

= a 32a2b + ab2ba2 + 2ab2b3

= a3 - 2a2b - ba2 + ab2 + 2ab2 - b3

= a33a2b + 3ab 2b3

= RHS

(ab)3 = a3 – 3a2b + 3ab2b3

Hence, we proved.

a2b = ba2

Multiplication is commutative

III. (x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

We know that the identity (x + a)(x + b) = x2 + (a + b)x + ab . Let us multiply this by a binomial ( x + c) . Then we get.

(x + a)(x + b)(x + c ) = [( x + a)(x + b)](x + c)

= ( x2 + (a + b)x + ab) × ( x + c)

= x [x2 + (a + b)x + ab] + c[ x2 + (a + b)x + ab ] (distributive law)

= x3 + (a + b )x2 + abx + cx 2 + (a + b)xc + abc

= x3 + ax 2 + bx 2 + abx + cx 2 + acx + bcx + abc

= x3 + (a + b + c)x2 + (ab + bc + ca )x + abc (Combine x2 , x terms)

(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc

Thus, we summarise the cubic identities as :

(a + b)3 = a 3 + 3a2b + 3ab2 + b3

(a b)3 = a 3 3a2b + 3ab2 b3

(x + a)(x + b)(x + c) = x 3 + (a + b + c)x2 + (ab + bc + ca )x + abc

Deductions:

The above identities give the following deductions:

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