# Question Bank - Equations - Set 2

• Number of Questions : 30
Topic : Theory of Equations
Solved ? : Yes
Source :

• Q1) If a + b + c = 0 , find the value of a^2/bc + b^2/ca + c^2/ab ?

• first method – find lcm of denominator i.e ( a^3+b^3+c^3)/abc = 3abc/abc= 3
Since a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)

Secod method- put value of a,b,c such that rhs should be zero as
a = 1 , b = 2 and c = -3
put these value in the expression u will get required ans

• Q2) If a + 1/a = 2 find a^37 + 1/a^100 ?

• Put a = 1 and rhs is satisfied so u will get ans as 2

• Q3) If (a - 1)^2 + (b + 2)^2 + (c + 1)^2 = 0 , then the value of 2a – 3b + 7c ?

• Sum of perfect squares should be zero if every term individually becomes zero .so
a - 1 = 0 => a = 1
b + 2 = 0 => b = -2
c + 1 = 0 => c = -1
so 2 * (1) – 3(-2) + 7(-1) = 2 + 6 - 7 = 1

• Q4) Minimum value of x^2 + 1/(x^2 + 1) - 3 ?

• Minimum value of perfect square should be zero .. so put x =0
You will get ans -2

• Q5) If x = 3 + 2√2 , Find √x + 1/√x ?

• As (a+b)^2=a^2 + b^2 + 2ab ,
Now equate 2ab = 2√2
ab = √2 x 1
so a= 1 and b = √2 , now
√x = (1+√2)
1/√x = √2-1

• Q6) If x^3 + y^3 = 35 , and x + y = 5 , find 1/x + 1/y .

• Given that x+y = 5 , cubing both sides
x^3 + y^3 + 3xy (x + y) = 125
35 + 15 xy = 125
xy = 6
now x+y/xy = 5/6
1/x + 1/y = 5/6

#ALTER
Put x = 3 and y = 2 … equations satisfied so
1/3 + 1/2 = 5/6

• Q7) If a = 70, b = 71 , c = -141 , find a^3 + b^3 + c^3 ?

• Here a + b + c = 0, using identity
a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)
a^3 +b^3 +c^3 = 3abc ( since a+b+c=0)
so ans -3 x 70 x 71 x 141

• Q8) If p = 124 , find (p(p^2 + 3p + 3) + 1)^1/3 ?

• First of all we open a bracket and try to it a identity so ..
p(p^2 + 3p + 3) + 1 = p^3 + 3p^2 + 3P + 1 = (p + 1)^3
Its cube root will be = (p + 1)
And p= 124
Ans = 125

#Alter
For better approach you have to take p=1 and solve u will get 8 and cube root of 8 is 2
It means if we take a value of p =1 , we get 2 , similarly if p = 125 we will get p = 125
No need so solve further for identity

• Q9) The unit digit of 27^16 – 18^48 ?

• This question seems easier but it is little tricky, now solve
Unit digit of any number can be find out by cyclicity.
cycle of unit digit is four so expression reduces into 7^4k – 8^4k = …..1 - ….6 = (-5) but unit digit can not in negatives so we take carry as 10 - 5 = 5

• Q10) If a/b = c/d = e/f = 3/1 , then (2a^2 + 4e^2 + 3c^2)/(2b^2 + 3d^2 + 4f^2) ?

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