In this section, we will discuss about what actually are algebraic expressions? How are they written? I know a lot must be going in your minds right now. Don’t you worry I’m here for you?

Basically, algebraic expressions are made up of integers constant, variables, and algebraic operations.

For example: 5x² + 3xy – 2 , 6x³ – 2x + 12 , etc.

Polynomial: An algebraic expression in which the variables involved have non – negative integral powers is called a polynomial.

P(x) = a_o + a₁ x+ a₂ x² + a₃ x³ +…

This was the general form of polynomials.

Like we just saw in the examples above 5x² + 3xy – 2, 6x³ – 2x + 12 are both polynomials.

The positive integral powers may go up to infinity. For example:  967x¹⁰⁰⁰⁰⁰ + 499 x ⁹⁹⁹⁹⁰ – 2000 is also a polynomial.

Que:- If (2 + √3)a = (2 – √3)b = 1 then what is the value of

             1/a + 1/b?

Solution:- given:- (2 + √3)a = (2 – √3)b = 1

Step 1:- First we will take the first term equals to 1 i.e.,

                                    (2 + √3)a = 1

We have from here,

            1/a = 2 + √3 ——-(i)

Step 2:- Taking the second term equals to one to obtain 1/b

     (2 – √3)b = 1

• 1/b = 2 – √3 ——–(ii)

As we now have the values of 1/a and 1/b, what the questions asked us was – 1/a + 1/b, so we are going to simply substitute the values from (i) and (ii) 

So, 1/a + 1/b = (2 + √3) + (2 – √3)        = 4

And that’s where you get your answer. All you need to do is just see what question is asking and go step by step to the solution. 

Que:- If 2p/ p² – 2p + 1= ¼ then find the value of (p + 1/p). 

Explanation- The approach is the same as that of the previous question, we will just go step by step.

Step 1:- All we need to do is find p + 1/p, we see that we don’t have 1/p, we will first try to make it in our equation this can be done when we divide numerator and denominator by p on the right-hand side of the equation.

Step 2:- As we have to find the value of p + 1/p

Therefore, 2/p-2+1p = ¼

p – 2 + 1/p = 8

p + 1/p = 10.

Que : If 5^5x+5 = 1, then find the value of x?

Approach I –

5^5x+5 = 1

• 5^5x+5 = 5^5x. 5⁵ = 1
• 5^5x = 1/ 5⁵
• 5^5x = 5^-5

Comparing both sides we get,

5x = -5

• x = -1

Approach 2:-  

5^5x+5 = 1

• 5^5x+5 = 5⁰               (since, 5⁰ = 1)
• 5x + 5 = 0
•  5x = -5
• x = -1

Que: If √x = √3 – √5 , then find the value of x² – 16x +6.  

 Solution:-

√x = √3 – √5

Squaring both sides,

x = 3 + 5 – 2√15

x – 8 =- 2√15

 squaring again,

x2 – 16x + 64 = 60

⇒x2 – 16x + 4 = 0

⇒x2 – 16x + 6 = 2

 Que: If n = 7 + 4√3, then find the value of √n + 1/√n ?

Solution:-

n = 7 + 4√3

n = 7 + 2 × 2√3  

n = 4 + 3 + 2 × 2√3   = ( 2 +√3)2

Therefore, √n = 2 + √3

Substituting the value of √3 in √n + 1/√n

√n + 1/√n = 2 +√3 + 1/2+√3

Que :- If 2s = a + b + c, then find the value of s(s – c) + (s – a) (s – b) ?

The approach that we are going to follow will be same i.e., we will just go as what the question is asking us and then go step by step to the solution.

Solution:- We have to find the value of s(s – c) + (s – a) (s – b), will go accordingly.

Therefore,  s(s – c) + (s – a) (s – b)

 = ¼ ( a + b+ c) (a + b – c) + ¼ (b + c –a )( a – b + c)

= ¼ [(a + b + c) (a + b – c) + (b + c – a) (a + c – b)]

= ¼ [(a + b)² – c² + ab + ac – a² + bc + c² – ac – b² – bc + ab]

= ¼  (a² + b² + 2ab – c² + ab + ac – a² + bc + c² – ac – b² – bc + ab)

= ¼  × 4ab = ab

 Q: If a, b, c are real numbers and a² + b² + c² = 2 (a – b – c) – 3, then the value of a + b + c is?

Solution:- we have been given that,

a2 + b2 + c2 = 2 (a – b – c) – 3

a2 + b2 + c2 = 2a – 2b – 2c – 3

We will try to put the similar terms together with a constant to make perfect square.

a2 – 2a +1 + b2 + 2b +1 + c2 + 2c +1  = 0

( a – 1)2 + ( b + 1)2 + ( c + 1)2 = 0

a – 1 = 0 , b + 1 =0 , c + 1 =0

a = 1, b=-1, c =-1

therefore, a + b + c = 1 – 1 – 1 = -1.

Let us now look at the algebraic identities.

Algebraic identities

1. (a + b)² = a² + 2ab + b²

2. (a – b)²  = a² – 2ab + b²

3. (a + b) ² = (a – b) ² + 4ab

4. (a – b) ²  = (a + b) ² – 4ab

 5. a² – b² = (a + b) (a – b)

6. a³ + b³ = (a + b) (a² – ab + b²)

 7. a³ – b³ = (a – b) (a² + ab + b²)

8. (a + b) ³ = a³ + b³ + 3ab (a + b)

9. (a – b) ³ = a³ – b³ – 3ab (a – b)

10. a³ + b³ = (a + b) ³ – 3ab (a + b)

11. a³ – b³ = (a – b) ³ + 3ab (a – b)

12. a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ac)

                       = (a + b + c) ½ (2a² + 2b² + 2c² – 2ab –2bc – 2ac)

                         = ½ (a + b + c) [(a – b)² + (b – c)² + (c – a)²]

13. If a + b + c = 0, then a³ + b³ + c³ = 3abc

14. (a + b + c) ³ = a³ + b³ + c³ + 3(b + c) (c + a) (a + b)

 15. a² + b² = (a + b)² – 2ab

16. a² + b² = (a – b) ² + 2ab

 17. (a + b + c) ² = a² + b² + c² + 2ab + 2ac + 2bc

18. a⁴ + b⁴ + a² b² = (a² – ab + b²) (a² + ab + b²)

Memorizing them will take time but you practice with questions they become a lot easier. Before moving ahead there are some rules, which will help to understand the topic better.

Rule1 :- a³ + 1/ a³  = (a + 1/a)³ – 3 ( a + 1/a)

Rule2 :- a³ – 1/ a³  = (a – 1/a)³ – 3 ( a – 1/a)

Rule3:- a + 1/a = 2 then aⁿ + 1/aⁿ = 2

Rule4:- a + 1/a = 2 then aⁿ – 1/aⁿ = 0

Rule5:- a + 1/a = 2 then a^m + 1/aⁿ = 2 (m≠n)

Rule6:- a + 1/a = -2 then aⁿ + 1/aⁿ = 2 if n is even and

              aⁿ + 1/aⁿ = -2 if n is odd

Rule7:-a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)

                                         = ½ (a + b + c) [(a – b)² + (b – c)² + (c –a)²]

Rule8:- If a + b+ c = 0, then a³ + b³ + c³ =3abc

Rule9:-  If a³ + b³ + c³ =3abc then either a + b+ c = 0 or a = b = c.

Rule10:- If a² + b² + c² = ab + bc + ca, then a = b = c.

 These are a few important rules which will help you with your practice. 

Que1: If (x + 1/x)² = 3, then find the value of x²⁰⁶ + x²⁰⁰ + x⁹⁰ + x⁸⁴ + x¹⁸ + x¹² + x⁶ + 1 ?

Approach:- Just look at the question carefully. What do you see? The value which we have to find has a common divisor or multiplier. This is vigilance is needed in such types of questions. Here, we have common multiplier 6, just analyse it and try to find the value of the least polynomial here i.e. x⁶.

We start by looking at what is given to us.

We have,

(x + 1/x)² = 3

• x + 1/x =√3 —————–(i)

On cubing both sides,

• (x + 1/x)³ = (√3 )³
• x³ + 1/ x³ + 3(x + 1/x) = 3√3
• x³ + 1/ x³ + 3√3= 3√3 (from (i))
• x³ + 1/ x³ = 0
• x⁶ + 1 =0

Therefore,

 x²⁰⁶ + x²⁰⁰ + x⁹⁰ + x⁸⁴ + x¹⁸ + x¹² + x⁶ + 1

= x²⁰⁰( x⁶ + 1) + x⁸⁴(x⁶ + 1) + x¹²(x⁶ + 1) + x⁶ + 1

= 0                                            (since,x⁶ + 1 =0)

 Que2: If a + b + c = 6, a² + b² + c² = 14 and a³ + b³ + c³ = 36, then find the value of abc ?

Approach:-1. As here we have been given the value of a + b+ c and a² + b² + c², the first thing to click your mind should be the formula of (a + b + c)².

(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

         (6)²     = 14 + 2(ab + bc + ca)

        36 = 14 + 2(ab + bc + ca)

• ab + bc + ca = 11

 2. The second thing that should strike your mind after looking at the question is the formula for (a + b + c)³.

From our list of formulas, look at the formula no. 12,

a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ac)

• 36 – 3abc = 6 ( 14 – 11)
• 36 – 3 abc = 18
•  36 – 18 = 3abc
• abc = 6

Que3: If x =³√5+ 2, then the value of x³ – 6x² + 12x – 13 is?

Solution:- Given that:- x =³√5+ 2

x – 2 =³√5

cubing on both sides, we get

(x – 2)³ = 5

x³ – 3x²  × 2 + 3x (-2)² – 2³ = 5

• x³ – 6x² + 12x – 8 = 5
• x³ – 6x² + 12x – 13 = 0
•  

Que4:  If a² + b² + c² = ab + bc + ac then find the value of  a+c/b?

Solution:- Since,a² + b² + c² = ab + bc + ca

• a² + b² + c² – ab – bc – ca = 0

On multiplying by 2,

• 2a² + 2b² + 2c² – 2ab – 2bc – 2ca = 0
•  a² + b² – 2ab + b² + c² – 2bc + c² + a² – 2ac = 0
•   (a – b)² + (b – c)² + (c – a)² = 0
•  a – b = 0  =>  a = b ,
•  b – c = 0  => b = c
•  c – a = 0 =>  c = a

 Therefore, a + c/b = 2a/a = a 

Que: If (3x – 2y) : (2x + 3y) = 5 : 6, then one of the values . start by taking cube root on both the sides of the equation