Cubic Identities: Numerical Example solved problems

__Application
of Cubic Identities__

** **

__I.____ Using the identity (a + b)^{3} = a ^{3} + 3a^{2} b + 3ab ^{2} + b^{3}__

** **

__Example 3.13__

Expand (*x* + 4)^{3}

*Solution:*

Comparing
(*x* + 4)^{3} with (*a* + *b*)^{3}
, we get *a* =
*x* , *b* = 4

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

(*x* + 4)^{3} =
(*x*)^{3} +
3(*x* )^{2} (4) +
3(*x*)(4)^{2} +
(4)^{3 }(replacing *a*, *b* values)^{}

= ( *x* )^{3} + 3*x*^{2} (4) + 3(*x*)(16) + 64

(*x* + 4)^{3} =
*x*^{3} +
12*x* ^{2} +
48*x* + 64

(4)^{2} = 4 × 4 = 16

(4)^{3} = 4 × 4 ×4=64

Try to expand this by using

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

** **

__Example 3.14__

Find the
value of (103)^{3}

*Solution:*

Now, (103)^{3}
=
(100 +
3)^{3}

Comparing
this with ( *a* +
*b*)^{3} , we get *a* = 100 , *b* =
3

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

(100 +
3)^{3 }= (100)^{3} + 3(100)^{2} (3) + 3(100)(3)^{2} + (3)^{3}

= 1000000+
3(10000)(3) + 3(100)(9) + 27

= 1000000+
90000 +
2700 +
27

(103)^{3
}=1092727

** **

__II.____ Using the identity (a - b)^{3} = a ^{3} - 3a^{2}
b + 3ab ^{2} - b^{3}__

** **

__Example 3.15__

Expand: (
*y* − 5)^{3}

*Solution:*

Comparing
( *y* − 5)^{3 }with (*a* − *b*)^{3}
, we get *a* =
*y* , *b* = 5

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

(*y *–5)^{3} = ( *y*)^{3} – 3( *y*)^{2} (5) + 3( *y*)(5)^{2}
- (5)^{3}

= ( *y*)^{3}
– 3*y*^{2} (5) + 3( *y*)(25) – 125

(*y*–5)^{3 }= *y*^{3} −15*y*^{2} + 75*y* −125^{}

Try to expand this by using

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

** **

__Example 3.16__

Find the
value of (98)^{3}

*Solution:*

Now, (98)^{3}
=
(100 −
2)^{3}

Comparing
this with ( *a* −
*b*)^{3} , we get *a* = 100 , *b* =
2

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

(100 −
2)^{3} = (100)^{3} −
3(100)^{2} (2) + 3(100)(2)^{2} – (2)^{3}

= 1000000
−
3(10000)(2) + 3(100)(4) – 8

= 1000000
−
60000 +1200
–
8

(98)^{3}
=
941192

** **

__III. Using
the identity ( x + a)(x
+ b )(x + c) = x^{3} +(a + b + c)x^{2}
+(ab + bc + ca )x + abc
__

** **

__Example 3.17__

Expand: (*x* + 3)(*x* + 5)(*x* +
2)

*Solution:*

Given

(*x* + 3)(*x* + 5)(*x* +
2)

Comparing
this with (*x* +
*a*)(*x* + *b*)(*x* + *c*)
, we get *x* =
*x* , *a* = 3 , *b* =
5 , *c* = 2

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

(*x* + 3)(*x* + 5)(*x* +
2) =
( *x* )^{3} +
(3 +
5 +
2)(*x*)^{2} +
(3 ×
5 +
5 ×
2 +
2 ×
3)*x* + (3)(5)(2)

*= x *^{3}* *+* *10*x*^{2}* *+* *(15* *+* *10* *+* *6)*x *+* *30

(*x* + 3)(*x* + 5)(*x* +
2) =
*x*^{3} +
10*x*^{2} +
31*x* + 30

** **

**Try these**

**Expand **: (i)** **(**
***x*** **+** **5)^{3} (ii) ( *y*
− 2)^{3} (iii) (*x* + 1)(*x* + 4)(*x* + 6)

**Solution:**

**(i)** Comparing (*x* + 5)^{3}
with (*a + b*)^{3}, we have *a* = *x* and *b* = 4.

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

(*x* + 5)^{3} = *x*^{3} + 3*x*^{2}
(5) + 3 (*x*) (5)^{2} + 5^{3}

. = *x*^{3} + 15*x*^{2} + 75*x *+
125

**(ii)** Comparing (*y* − 2)^{3}
with (*a − b*)^{3} we have *a* = *y* *b* = *z*

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

(*y* − 2)^{2} = *y*^{3} – 3*y*^{2}(2)
+ 3*y*(2)^{2} + 2^{3}

= *y*^{3} − 6*y*^{2} + 12*y* + 8

**(iii)** Comparing (*x* + 1) (*x*
+ 4) (*x* + 6) with (*x* + *a*) (*x + b*) (*x + c*) we
have

*a* = 1 *b* = 4 and *c* = 6

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

= *x*^{3} + (1 + 4 + 6) *x*^{2} + ((1)
(4) + (4) (6) + (6) (l))*x* + (1) (4) (6)

= *x*^{3} + 11*x*^{2} + (4 + 24 + 6) *x*
+24

= *x*^{3} + 11*x*^{2} + 34*x* +
24

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8th Maths : Chapter 3 : Algebra : Application of Cubic Identities | Questions with Answers, Solution | Algebra | Chapter 3 | 8th Maths

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