ALGEBRAIC IDENTITIES

   

ALGEBRAIC IDENTITIES

  An Algebraic Identity is an algebraic equation which is true for all values of variables in them.

For example,the algebraic identity,

(a+b)(a - b)= a² - b²,is always true for all values of ( a ) and (b).

Some of the common,most important ones are: 

(ⁱ)( a+b)²=
                   (a+b)(a+b)
               = a(a+b)+b(a+b)
               =a²+ab+ab+b²
               =a²+2ab+b²

*(a+b)² =a²+2ab+b²

  Example 1

Find (3x+2y)², using algebraic identities.

Solution:

(3x+2y)²

Using (a+b)² = a²+2ab +b²

a=3x,b=2y

            = (3x)²+2(3x)(2y)+(2y)²
            = 9x²+12xy+4y²

{the square affects all characters in the bracket}

Hope you understand!

Example 2

Expand:  (x+y)²

a=x,b=y

=Using (a+b)² = a²+2ab+b²,we have:

(x+y)² = x² +2xy + y²

B.(a+b)(a-b)

by expansion,we have:

       a (a-b)+b(a-b)
    = a²-ab+ab-b²
    = a² - b²

*(a+b)(a-b) = a²-b²

Example 3

Evaluate 297×303,using algebraic identities.

Solution:

Using (a+b)(a-b) = a² -b²

       297×303
  =   (300 - 3)(300+3)
          a =300,b=3
  =   a² - b²
  =    300² - 3²
  =    90000 - 9
  =    89991.

C.(a-b)² = a²-2ab+b²

Example 4

Expand (1-x)²

solution:

a=1,b=x

  =1² -2×1×x+x²

  =1 -2x+x².

Example 5.

Evaluate (197)²

Using (a - b)² = a² -2ab +b²,

(197)² can be written as :

(200 - 3)²

            ( a =200,b=3)

= 200² -2×200×3 + 3²
= 40000 -1200+9
= 38809

D.(a+b)(a² -ab+b²)

 =a³ -a²b+ab² +a²b - ab² +b³= a³+b³

*a³+b³ =  (a+b)(a² - ab+b²)

  Example 6

  show that a³+b³ = (a+b)³-3ab(a+b)

solution:

(a+b)³ = (a+b)(a+b)(a+b)
            = (a+b)(a+b)²
            = (a+b)(a²+2ab+b²)
            = a³+2a²b+ab²+a²b +2ab² +b³
            = a³+3a²b+3ab²+b³
            = a³+3ab(a+b)+b³ (factorised)

            = a³+b³+3ab(a+b)

Writing out the equation,we have:

(a+b)³  = a³+b³+3ab(a+b)

Make a³+b³ the subject of formula.

      a³+b³  = (a+b)³ -3ab(a+b).

E.   (a -b)(a²+ab+b²)

     = a³+a²b+ab² -a²b -ab² -b³

   *a³ -b³ = (a-b)(a²+ab+b²)

Example 7

Show that:
        a³ - b³ = (a-b)³ -3ab(a-b)

Solution:

     (a -b)³ = (a-b)(a-b)(a-b)
                 = (a-b)(a-b)²
                 = (a-b)(a² -2ab+b²)
                 = a³ -2a²b+ab²-a²b+2ab² -b³

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

Making (a³ -b³) the subject of formula,

        a³ - b³  = (a - b)³ + 3ab(a - b).



F.  (x+a)(x+b) = x²+bx+ax+ab
                         = x² +x(a+b)+ab.

*(x+a)(x+b) =x² +x(a+b)+ab

Example 8 

Expand
(yz+3)(yz - 5)

Solution:

Using the identity

(x+a)(x +b) = x² +x(a+b)+ab,

Where x =yz,a =3 and b =-5,

   (yz+3)(yz-5)
= (yz)²+yz(3+-5)+3(-5
= y²z²+yz(-2)-15
= y²z² -2yz -15


G. (a+b+c)² = a²+b²+c²+2ab+2ac+2bc

                    = a²+b²+c²+2(ab+ac+bc).

   *(a+b+c)² = a²+b²+c²+2(ab+ac+bc)

Example 9

If a²+b²+c²  = 14, and a+b+c = 6,
Find the value of  (ab+ac+bc).

Solution:

From :

  (a+b+c) = a²+b²+c²+2(ab+ac+bc)

                  6²  =  14 + 2(ab+ac+bc)
                 36   =  14 + 2(ab+ac+bc)
  2(ab+ac+bc) = 36-14
  2(ab+ac+bc) = 22

       ab+ac+bc =22÷2
   ab + ac + bc =11.

Example 10.

If x+y =12 and xy = 32, Find x²+y²

Solution:

x²+y²  = (x+y)² -2xy.

{from (a+b)² = a² +2ab +b² }

=12² -2(32)
=144-64
=80

Example 11

If a+ (1/a ) = 5,find find the value of
a³+(1/a³)

Solution:

From

  (a+b)³ =  a³ +b³ +3ab(a+b),

We have:

{a+(1/a)}³ = a³ +(1/a³) + 3×a× 1/a{a+(1/a)}

                  = a³+ (1/a³) + 3{a+(1/a)}

Putting the value of {a+(1/a)} = 5,

We have:

           5³    = a³ +(1/a)³ + 3×5
        125    = a³ +(1/a)³+15
a³ + (1/a)³ = 125 -15
                  = 110

See Series