RD Chapter 5 Factorisation of Algebraic Expressions Ex VSAQS Solutions
Question - 1 : - If a + b + c = 0, then write the value of a3 + b2 + c2.
Answer - 1 : -
∵ a + b+ c =0, ‘
Then a3 + b2 + c3 = 3 abc
Question - 2 : - If a2 + b2 + c2 = 20 and a + b + c = 0, find ab + bc + ca.
Answer - 2 : -
a2 +b2 + c2 = 20, a + b + c = 0
∴ (a + b+ c)2 = 0
a2 + b2 + c2 + 2(ab + bc + ca)= 0
⇒ 20 + 2(ab + be + ca) = 0
⇒ 2(ab + bc + ca) = -20
ab + bc + ca = −202 =-10
Question - 3 : - If a + b + c = 9 and ab + bc + ca = 40, find a2 + b2 + c2.
Answer - 3 : -
a + b + c = 9, ab + bc + ca = 40
Squaring both sides,
(a + b + c)2 = (9)2
a2 + b2 + c2 + 2{ab + bc+ ca) = 81
⇒ a2 +b2 + c2 + 2 x 40 = 81
⇒ a2 + b2 + c2 + 80 = 81
a2 + b2 + c2 = 81 – 80 = 1
Question - 4 : - If a2 + b2 + c2 = 250 and ab + bc + ca = 3, find a + b + c.
Answer - 4 : -
a2 + b2 + c2 = 250, ab + bc +ca = 3
(a+ b + c)2 = a2 + b2 + c2 +2(ab + bc +ca)
= 250+ 2 x3 =250 + 6 = 256
=(±16)2
∴ a + b + c = ±16
Question - 5 : - Write the value of 253 – 753 + 503.
Answer - 5 : -
253 – 753 + 503
Let a = 25, b = -75 and c = 50
∵ a + b + c = 25 – 75 + 50 = 0
∴ a3 +b2 + c3 = 3abc
⇒ 253 + (-75)3 + 503
= 3 x 25 x (-75)- x 50 = -281250
Question - 6 : - Write the value of 483 – 303 – 183.
Answer - 6 : -
483 – 303 – 183
Let a = 48, b = -30, c = -18
∵ a + b+ c = 48 – 30 – 18 = 0
∴ a2 + b2 + c2 = 3abc
⇒ 483 – 303 – 183
= 3 x 48 x (-30) (-18)
= 77760
Question - 7 : - 
Answer - 7 : -
Question - 8 : - Write the value of 303 + 203 – 503.
Answer - 8 : -
303 + 203 – 503
Let a = 30, b – 20, c = -50
∵ a + b + c = 30 + 20 -50 = 50 – 50 = 0
∴ a3 +b3 + c3 = 3 abc
⇒ 303 + 203 – 503 = 3 x 30 x 20 x(-50)
= 90000
Question - 9 : - Factorize: x4 + x2 + 25.
Answer - 9 : -
x4 + x2 + 25
⇒ (x2)2 + (5)2 + 2x2 x5 – 2x2 x 5 +x2
⇒ (x2)2 + (5)2 + 10x2 –10x2 + x2
= (x2)2 + (5)2 + 10x2 –9x2
= (x2 + 5)2 – (3x)2 {∵ a2-b2 =(a + b) (a – b)}
= (x2 + 5 – 3x) (x2 + 5 + 3x)
= (x2 – 3x + 5) (x2 + 3x + 5)
Question - 10 : - Factorize: x2 – 1 – 2a – a2.
Answer - 10 : -
x2 – 1 – 2 a – a2
= x2 – (1 +2a + a2) – (x)2 –(1 + a)2 {∵ a2 –b2 = (a + b) (a – b)}
= (x + 1 + a) (x – 1 – a)
= (x + a + 1) (x – a – 1)