An identity is an equality that remains true regardless of the values chosen for its variables.

**Algebraic Identities**

An identity is an equality that remains true regardless of the values
chosen for its variables.

We
have already learnt about the following identities:

1.
(*a* + *b* )^{2 }≡ *a* ^{2} + 2*ab* +*b*^{2}

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

3.
(*a* + *b* )(*a* −*b*) ≡ *a* ^{2} −*b*^{2}

4.
(*x* + *a* )(*x* +*b*) ≡ *x* ^{2} +(*a*
+
*b* )*x* +*ab*

**Note**

(i) a^{2} +* b*^{2}
= (a +b)^{2} −2ab

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

**Example 3.16**

Expand
the following using identities: (i) (3*x* + 4*y*)^{2 }(ii) (2*a*
− 3*b*)^{2 }(iii) (5*x* + 4*y*)(5*x* −
4*y*)^{ }(iv) (*m* + 5)(*m* −
8)^{}

*Solution*

(i) (3*x
+* 4y)^{2} [we have
(a + b)^{2 } = *a*^{2} + 2ab +* b*^{2} ]

(3*x +*
4*y*)^{2} = (3*x*)^{2} + 2(3*x*)(4*y*) +(4*y*)^{2} [put [a = 3*x*, * b *= 4*y*]

= 9*x*^{2} + 24*xy* +16*y*^{2}

(ii)
(2*a* − 3*b*)^{2}

[ we have (a − b)^{2} = *a*^{2} + 2*ab* +* b*^{2}* *]

(2*a* − 3*b*)^{2}
= (2a)^{2} −2(2*a*)(3*b*) +(3*b*)^{2} put [a = 2*a*,* b *= 3*b*]

= 4*a* ^{2} −12*ab* + 9*b*^{2}

(iii)
(5*x +* 4*y*)(5*x* − 4*y*)

[ we have (a + b)(a −b) = *a*^{2}
–*b*^{2}]

(5*x +*
4*y*)(5*x* − 4*y*) = (5*x*)^{2} −(4*y*)^{2} put [a = 5*x b *= 4*y*]

= 25x^{2} −16*y*^{2}

(iv) (*m*
+ 5) (*m* − 8)

[we have (*x* + *a*)(*x*-*b*) = *x*^{2}
+ (*a-b*)*x*-*ab*]

(*m* + 5) (*m* − 8) = *m*^{2} +(5
− 8)*m* −(5)(8) put [*x* = *m*, *a* = 5, *b* = 8]

= *m*^{2}
− 3*m* – 40

** **

We know that
(*x* + *y*)^{2} = *x* ^{2} +
2*xy* + *y*^{2}

Put *x*
=
*a* +*b*, *y* = *c*

Then, (*a* +
*b* +*c*)^{2} = (*a* +*b* )^{2} +
2(*a* +*b*)(*c* ) +*c*^{2}

*=
a *^{2}* *+* *2*ab *+*
b*^{2}* *+* *2*ac *+*
*2*bc *+*c*^{2}

*=
a*^{2}* *+*b*^{2}* *+* c *^{2}* *+* *2*ab *+*
*2*bc *+*
*2*ca*

Thus, (*a* + *b* +*c* )^{2} ≡ *a* ^{2}
+*b* ^{2} + *c* ^{2}
+ 2*ab* + 2*bc* + 2*ca*

**Example 3.17**

Expand
(*a* − *b* +*c*)^{2}

*Solution*

Replacing
‘*b*’ by ‘-*b* ’ in the expansion of

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

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

*=
a *^{2}* *+*b*^{2}* *+* c *^{2}* *−2*ab *−*
*2*bc *+*
*2*ca*

**Progress Check**

Expand the following and verify :

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

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

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

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

**Example 3.18**

Expand
(2*x* + 3*y* + 4*z*)^{2}

*Solution*

We
know that,

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

Substituting,
*a* = 2*x* , *b* = 3*y* and*c* =
4*z*

(2*x*
+
3*y* + 4*z*)^{2} = (2*x* )^{2} +(3*y*)^{2} + (4*z* )^{2} + 2(2*x* )(3*y* ) + 2(3*y*)(4*z* ) + 2(4*z* )(2*x*)

= 4*x* ^{2} + 9*y*^{2} + 16*z* ^{2} +12*xy* + 24*yz* +16*xz*

**Example 3.19**

Find
the area of square whose side length is 3*m* +
2*n* − 4*l*

*Solution*

Area
of square = side × side

= (3*m* + 2*n* − 4*l*)×(3*m* + 2*n* − 4*l*)

= (3*m* + 2*n* − 4*l*)^{2}

We
know that, (*a* + *b* +*c*)^{2} =
*a*^{2} + *b* ^{2} +*c*^{2} + 2*ab* +
2*bc* + 2*ca*

[
3*m* + 2*n* + (−4*l*)]^{2 }= (3*m*)^{2} +(2*n*)^{2} + (−4*l*)^{2} + 2(3*m*)(2*n*) + 2(2*n*)(− 4*l* ) + 2(−4*l*)(3*m*)

= 9*m*^{2} + 4*n*^{2} + 16*l*^{2} +12*mn* − 16*ln* −24*lm*

Therefore,
Area of square = [9*m*^{2} +
4*n*^{2} + 16*l*^{2} +12*mn*
−
16*ln* −24*lm*] sq.units.

**Substituting**

*a *= 3*m,*

*b *= 2*n*

*c *= –4*l*

** **

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

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

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

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

**Example 3.20**

Expand
the following:

(*i*)
(*x* + 5)(*x* + 6)(*x* + 4) (*ii* ) (3*x* − 1)(3*x*
+ 2)(3*x* − 4)

*Solution*

We
know that (*x*
+
*a* )(*x* +*b*)(*x* +*c*) = *x* ^{3} +(*a*
+
*b* +*c*)*x* ^{2} + (*ab* +*bc* + *ca*)*x* +*abc* --(1)

(i)
(*x* + 5)(*x* + 6)(*x* +
4)

Replace: *a *by 5, *b *by 6, *c *by 4 in (1)

*=
x *^{3}* *+(5* *+*
*6* *+*
*4)*x*^{2}* *+*
*(30* *+*
*24* *+*
*20)*x *+(5)(6)(4)

*=
x *^{3}* *+15*x*^{2}* *+*
*74*x *+120

(ii)
(3*x* − 1)(3*x* + 2)(3*x* −
4)

Replace : * x* by 3*x*,
a by –1, *b* by 2,* c *by –4 in (1)

=
(3*x*)^{3} +(− 1 + 2 − 4)(3*x*)^{2} +(−2
−
8 +
4)(3*x*) + (−1)(2)(−4)

= 27*x*^{3}
+(− 3)9*x*^{2} +(− 6)(3*x*) + 8

= 27*x*^{3}
−27*x*^{2} − 18*x +* 8

** **

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

substituting
*a* = *b* = *c* =
*y* in the identity

we
get, (*x* + *y* )(*x* +
*y* )(*x* + *y*) =
*x*^{3} +(*y* +
*y* +*y*)*x*^{2} + (*yy* +*yy* + *yy* )*x* + *yyy*

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

Thus, (*x* + *y*)^{3 }≡ *x*^{3} + 3*x*^{2}*y* + 3*xy*^{2} +*y*^{3 }(or) (*x* + *y*)^{3 }≡ *x*^{3} +*y*^{3} + 3*xy*(*x* +*y*) by replacing *y*
by -*y*, we get

(*x* − *y* )^{3 }≡ *x*^{3} − 3*x*^{2}*y* + 3*xy*^{2} −*y*^{3 }(or) (*x* − *y* )^{3 }≡ *x*^{3} − *y*^{3} − 3*xy*(*x* −*y*)

**Example 3.21**

Expand
(5*a* − 3*b*)^{3}

*Solution*

We
know that,

(*x*
-*y*)^{3}= *x* ^{3} − 3*x*^{2}*y* + 3*xy* ^{2} −*y*^{3}

(5*a*
− 3*b*)^{3 }= (5*a* )^{3} − 3(5*a*)^{2}(3*b* ) + 3(5*a*)(3*b* )^{2} –
(3*b*)^{3}

=
125*a* ^{3} − 3(25*a*^{2} )(3*b*
) +
3(5*a*)(9*b* ^{2} ) – (3*b*)^{3}

= 125*a* ^{3} −225*a*^{2}*b* + 135*ab* ^{2} −27*b*^{3}

The
following identity is also used:

* x *^{3}* *+* y *^{3}* *+* z *^{3}* *−* *3*xyz *≡*
*(*x *+*
y *+*
z *)(*x*^{2}* *+*
y*^{2}* *+* z*^{2}* *−* xy *−*yz
*−*zx*)

We
can check this by performing the multiplication on the right hand side.

**Note**

(i) If (*x* + *y* + *z*) = 0 then *x* ^{3} + *y* ^{3} + *z* ^{3} = 3*xyz*

Some identities involving sum, difference and product are stated
without proof

(i) *x*^{3} + *y*^{3} ≡ (*x* + *y* )^{3}
− 3*xy*(*x* + *y*) (ii) *x*^{3} − *y*^{3} ≡ (*x* − *y* )^{3} + 3*xy*(*x* −*y*)

**Example 3.22**

Find
the product of (2*x* + 3*y* +
4*z* )(4*x*^{2} + 9*y* ^{2} +16*z*
^{2} − 6*xy* −12*yz* −
8*zx*)

*Solution*

We
know that, (*a* + *b* +*c* )(*a*^{2} +
*b* ^{2} +*c*^{2} − *ab* −*bc* −*ca*) = *a* ^{3} +*b*^{3} + *c* ^{3} −
3*abc*

(2*x*
+
3*y* + 4*z* )(4*x* ^{2} +
9*y* ^{2} +16*z* ^{2} −
6*xy* −12*yz* − 8*zx*)

= (2*x* )^{3} +(3*y*)^{3} + (4*z* )^{3} − 3(2*x* )(3*y* )(4*z*)

= 8*x* ^{3} + 27*y* ^{3} + 64*z* ^{3} −72*xyz*

**Example 3.23**

Evaluate
10^{3} − 15^{3} +
5^{3}

*Solution*

We
know that, if *a *+* b *+*c
*=*
*0* *, then* a *^{3}*
*+*
b *^{3}* *+*c*^{3}* *=* *3*abc*

Here,
*a* + *b* +*c *=10−15+5 = 0

Therefore,
10^{3} + (−15)^{3} +
5^{3 }= 3(10)(−15)(5)

10^{3}
−15^{3}
+5^{3
}= −2250

Replace *a* by 10,* b *by –15, c by 5

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