We shall prove it now.

**Cubic Identities**

** **

**I. ( a + b)^{3}
= a^{3} + 3a^{2}b + 3ab^{2} + b^{3}**

We shall
prove it now,

*LHS *= (*a* + *b*)^{3}

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

= (*a* + *b*)^{2} (*a* + *b*)

= (*a* ^{2} + 2*ab* + *b*^{2}
)(*a* + *b*) (using identity)

=* a*(*a*^{2}* *+* *2*ab
*+* b*^{2}* *)* *+* b*(*a *^{2}* *+* *2*ab
*+* b*^{2}* *)* *(using distributive law)

=* a *^{3}* *+* *2*a*^{2}*b *+* ab*^{2}* *+* ba*^{2}* *+* *2*ab*^{2}* *+* b*^{3}

=* a*^{3}* *+* *(2*a*^{2}*b *+* ba*^{2}* *)* *+* *(*ab*^{2}* *+* *2*ab*^{2}* *)*
*+* b*^{3 }(grouping ‘like’ terms)^{}

=* a *^{3}* *+* *3*a*^{2}*b *+* *3*ab *^{2}* *+* b*^{3}

= *RHS*

**( a + b)^{3} = a^{3} + 3a^{2}b + 3ab^{2} + b^{3}**

Hence, we proved the cubic identity by direct multiplication.

**Activity**

You can visualize
the geometrical proof of (*a*+*b*)^{3} with the help of* y*our teacher.

** **

**II. ( a – b)^{3}
= a^{3} – 3a^{2}b + 3ab^{2} – b^{3}**

We can prove
this identity by direct multiplication

We have (*a* − *b*)^{3}
=
(*a* − *b*)(*a* − *b*)(*a* − *b*)

= (*a* − *b*)^{2} × (*a* − *b*)

= (*a* ^{2} − 2*ab* + *b*^{2} )(*a* − *b*)

= *a*(*a*^{2} − 2*ab* + *b* ^{2} ) − *b*(*a* ^{2} − 2*ab* + *b*^{2} )

= *a* ^{3} − 2*a*^{2}*b* + *ab*^{2} − *ba*^{2} + 2*ab*^{2} − *b*^{3}

*= a*^{3}* - *2*a*^{2}*b - ba*^{2}* + ab*^{2}* + *2*ab*^{2}* - b*^{3}

= *a*^{3} − 3*a*^{2}*b* + 3*ab* ^{2} − *b*^{3}

=* RHS*

**( a – b)^{3} = a^{3} – 3a^{2}b + 3ab^{2} – b^{3}**

Hence, we
proved.

*a*^{2}*b *=* ba*^{2}

Multiplication is commutative

** **

**III. **(** x + a**)(

We know that
the identity (*x* +
*a*)(*x* + *b*) =
*x*^{2} +
(*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*)

= ( *x*^{2} +
(*a* + *b*)*x* + *ab*)
×
( *x* + *c*)

=* x *[*x*^{2}* *+* *(*a
*+* b*)*x *+* ab*]*
*+* c*[* x*^{2}* *+* *(*a
*+* b*)*x *+* ab* ] (distributive
law)

= *x*^{3}* *+* *(*a *+* b *)*x*^{2}* *+* abx
*+* cx *^{2}* *+* *(*a *+* b*)*xc
*+* abc*

= *x*^{3}* *+* ax *^{2}* *+* bx *^{2}* *+* abx *+* cx *^{2}* *+* acx *+* bcx
*+* abc*

= *x*^{3} + (*a* + *b* + *c*)*x*^{2} + (*ab* + *bc* + *ca* )*x* + *abc* (Combine *x*^{2} , *x* terms)

(** x + a**)(

Thus, we
summarise the cubic identities as :

**• ( a + b)^{3} = a ^{3}
+ 3a^{2}b
+ 3ab^{2} + b^{3}**

**• ( a − b)^{3} = a ^{3}
− 3a^{2}b
+ 3ab^{2} − b^{3}**

**• ( x + a)(x
+ b)(x
+ c) = x ^{3}
+ (a + b + c)x^{2}
+ (ab + bc + ca )x + abc**

**Deductions:**

The above
identities give the following deductions:

Tags : Algebra | Chapter 3 | 8th Maths , 8th Maths : Chapter 3 : Algebra

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