1. Find the cube with the help of the formulae:
1) 2x2 + 3y2
Solution: (2x2 + 3y2)3
= (2x2)3 + 3.(2x2)2.3y2 + 3.2x2.(3y2)2 + (3y2)3
= 8x6 + 3.4x4.3y2 + 3.2x2.9y4 + 27y6
= 8x6 + 36x4y2 + 54x2y4 + 27y6
2) 7m2 -2n
Solution: (7m2 – 2n)3
= (7m2)3 – 3.(7m2)2.2n + 3.7m2.(2n)2 – (2n)3
= 343m6 – 3.49m4.2n + 3.7m2.4n2 – 8n3
= 343m6 – 294m4n + 84m2n2– 8n3
3) 2a – b – 3c
Solution: (2a – b – 3c)3
= {(2a – b) – 3c}3
= (2a – b)3 – 3.(2a – b)2.3c + 3.(2a – b).(3c)2 – (3c)3
= (2a)3 – 3.(2a)2.b + 3.2a.b2 – b3– 3{(2a)2 – 2.2a.b + b2)}3c + (6a – 3b).9c2 – 27c3
= 8a3 – 3.4a2b + 6ab2 – b3 – 3(4a2 – 4ab + b2)3c + 54ac2 – 27bc2 – 27c3
= 8a3 – 12a2b + 6ab2 – b3 – 36a2c + 36abc – 9b2c + 54ac2 – 27bc2 – 27c3
2. Simplify:
1) (7x + 3b)3 – (5x + 3b)3 – 6x(7x + 3b)(5x + 3b)
Solution: Let, 7x + 3b = m, 5x + 3b = n
Now, m – n = 7x + 3b – 5x – 3b = 2x
Given expression = (m)3 – (n)3 – 6x.mn
= (m)3 – (n)3 – 3mn.2x
= (m)3 – (n)3 – 3mn(m – n)= (m – n)3
= (2x)3
= 8x3
2) (a + b + c)3 – (a – b – c)3 – 6(b + c){a2 – (b + c)2}
Solution: Let, a + b + c = a + (b + c) = m; a – b – c = a -(b + c) = n
Now, m – n = a + b + c – a + b + c = 2(b + c)
Given expression = (a + b + c)3 – (a – b – c)3– 6(b + c){a2 – (b + c)2}
= (a + b + c)3 – (a – b – c)3 – 3.2(b + c)(a + b + c)(a – b – c)
= m3 – n3 – 3.(m – n)m.n
= m3 – n3 – 3m2n + 3mn2
= (m – n)3
= {2(b + c)}3
= 8(b + c)3
3) (m + n)6 – (m – n)6 – 12mn(m2 – n2)2
Solution: (m + n)6 – (m – n)6 – 12mn(m2 – n2)2
= {(m + n)2}3 – {(m – n)2}3 – 3.4mn(m2 – n2)2
= {(m + n)2}3 – {(m – n)2}3 – 3.4mn{(m + n)(m – n)2
= {(m + n)2}3 – {(m – n)2}3 – 3.4mn{(m + n)2(m – n)2
Let, (m + n)2 = p; (m – n)2 = q
and, p – q = (m + n)2 – (m – n)2 = (m + n – m + n)(m + n + m – n) = 2n.2m = 4mn
Given expression = p3 – q3 – 3.(p – q)p.q
= p3 – q3 – 3pq(p – q)= (p – q)3
= {(m + n)2 – (m – n)2}3
= (4mn)3 [a + b)2– (a – b)2 = 4ab]
= 64m3n3
4) (x + y)(x2 – xy + y2) + (y + z)(y2 – yz + z2) + (z + x)(z2 – zx + x2)
Solution: (x + y)(x2 – xy + y2) + (y + z)(y2 – yz + z2) + (z + x)(z2 – zx + x2)
= (x3 + y3) + (y3 + z3) + (z3 + x3)
= x3 + y3 + y3 + z3 + z3+ x3
= 2x3 + 2y3 + 2z3
= 2(x3 + y3 + z3)
5) (2x + 3y – 4z)3 + (2x – 3y + 4z)3 + 12x{4x2 – (3y – 4z)2}
Solution: (2x + 3y -4z)3 + (2x – 3y + 4z)3 + 12x{4x2 – (3y – 4z)2}
= (2x + 3y – 4z)3 + (2x – 3y + 4z)3 + 12x{(2x)2 – (3y -4 z)2}
= {2x + (3y – 4z)}3 + {2x – (3y – 4z)}3 + 12x{2x – (3y – 4z)}{2x + (3y – 4z)}
={2x + (3y – 4z)}3 + {2x – (3y – 4z)}3 + 3.4x{2x – (3y – 4z)}{2x + (3y – 4z)}
Let, 2x + (3y – 4z) = m; 2x – (3y – 4z) = n
Now, m + n = 2x + 3y – 4z + 2x – 3y + 4z = 4x
Given expression = m3 + n3 + 3(m + n)mn
= m3 + n3 + 3mn(m + n)= (m + n)3
= (4x)3
= 64x3
3. If a – b = 5 and ab = 36 , what is the value of a3 – b3 ?
Solution: Given, a – b = 5 and ab = 36
We know, a3 – b3= (a – b)3 + 3ab(a – b)
= 53+ 3.36.5
= 25 + 540
= 665
4. If a3 – b3 = 513 and a – b = 3 , what is the value of ab?
Solution: Given, a3 – b3 = 513 and a – b =3
We know, a3 – b3 = (a – b)3 + 3ab(a – b)
Or, 513 = 32 + 3ab.3
Or, 513 = 27 + 9ab
Or, 9ab = 513 – 27
Or, 9ab = 486
Or, ab = 486/9
Or, ab = 54
5. If x = 19 and y = -12 , find the value of 8x3 + 36x2y + 54xy2 + 27y3 .
Solution: Given, x = 19 and y = -12
Given expression = 8x3 + 36x2y + 54xy2 + 27y3
= (2x)3 + 3.(2x)2 3y + 3.2x.(3y)2 + (3y)3
= (2x + 3y)3
= {2.19 + 3.(-12)}3
= (38 – 36)3
= (2)3
= 8
6. If a = 15, what is the value of 8a3 + 60a2 + 150a + 130 ?
Solution: Given, a = 15
Given expression =8a3 + 60a2 + 150a + 130
= (2a)3+ 3.(2a)2 .5 + 3.2a.52 + 53 + 5
= (2a + 5)3 + 5
= (2.15 + 5)3 + 5
= (30 + 5)3 + 5
= (35)3 + 5
= 42875 + 5
= 42880
7. If a + b = m, a2 + b2 = n and a3 + b3 = p3, show that, m3 + 2p3 = 3mn
Solution: Given, a + b = m, a2 + b2 = n and a3 + b3 = p3
L.H.S = m3 + 2p3
= (a + b)3 + 2.(a3 + b3) [Substituting value]
= a3 + 3a2b + 3ab2 + b3 + 2a3 + 2b2
= 3a3 + 3b3 + 3a2b + 3ab2
= 3(a3 + b3 + a2b + ab2)
= 3(a3 + a2b + b3 + ab2)
= 3{a2( a+ b) + b2(a + b)}
= 3.(a + b)(a2 + b2)
= 3.m.n [Substituting value]
= 3mn
= R.H.S (Proved)
8. If a + b = 3 and ab = 2, find the value of (a) a2 – ab + b2 and (b) a3 + b3 .
Solution: Given, a + b = 3 and ab = 2
(ক) a2 – ab + b2
For given expression (a), we have
a2 – ab + b2
= a2 + b2 – ab
= (a + b)2 – 2ab – ab [a2 + b2 = (a + b)2 – 2ab]
= 32 – 2.2 -2
= 9 – 4 – 2
= 9 – 6
= 3
For given expression (b), we have
a3 + b3
= (a + b)3 – 3ab(a + b)
= 33 – 3.2.3
= 27 – 18
= 9
9. If a – b = 5 and ab = 36, find the value of (a) a2 + ab + b2 and (b) a3 – b3 .
Solution: Given, a – b = 5 এবং ab = 36
For given expression (a), we have
a2 + ab + b2
= (a – b)2 + 2ab + ab [a2 + b2 = (a – b)2 + 2ab]
= 52 + 3ab
= 25 + 3.36
= 25 + 108
= 133
For given expression (b), we have
a3 – b3 = (a – b)3 + 3ab(a – b)
= 53 + 3.36.5 [Substituting value]
= 125 + 540
= 665
10. m + 1/m = a, find the value of m3 + 1/m3 .
Solution: Given, m + 1/m = a
Given expression = m3 + 1/m3
= (m + 1/m)3 – 3.m.1/m(m + 1/m)
= a3 – 3a [substituting value]
11. If x – 1/x = p, find the value of x3 – 1/x3 .
Solution: Given, x – 1/x = p
Given expression = x3 – 1/x3
= (a – 1/x)3 + 3.x.1/x(x – 1/x)
= (a – 1/x)3 + 3.(x – 1/x)
= p3– 3p [substituting value]
12. If a – 1/a = 1, show that, a3 – 1/a3 = 4
Solution: Given, a – 1/a = 1
L.H.S. = a3 – 1/a3
= (a – 1/a)3 + 3.a.1/a(a – 1/a)
= 13 + 3.1
= 1 + 3
= 4
= R.H.S. (proved)
13. If a + b + c = 0, show that, 1) a3 + b3 + c3 = 3abc
1) a3 + b3 + c3 = 3abc
Solution: Given,
a + b + c = 0
Or, a + b = -c
Or, (a + b)3 = (-c)3
Or, a3 + b3 + 3ab(a + b) = -c3
Or, a3 + b3 + 3ab(-c) = -c3 [a + b = -c]
Or, a3+b3– 3abc = -c3
Or, a3 + b3 + c3 = 3abc (showed)
14. If p – q = r, show that, p3 – q3 – r3 = 3pqr
Solution: Given, p – q = r
Or, (p – q)3 = r3
Or, p3 – q3 – 3pq(p – q) = r3
Or, p3 – q3 – 3pq.r = r3
Or, p3 – q3 – r3 = 3pqr (showed)
15. 2x – 2/x = 3, showed that, 8(x3 – 1/x3) = 63
Solution: Given, 2x – 2/x = 3
Or, 2(x – 1/x) = 3
Or, x – 1/x = 3/2
Or, (x – 1/x)3 = (3/2)3
Or, x3 – (1/x)3 – 3.x.1/x.(x – 1/x) = 27/8
Or, x3 – 1/x3 – 3.(x – 1/x) = 27/8
Or, x3 – 1/x3 – 3.(3/2) = 27/8
Or, x3 – 1/x3 – 9/2 = 27/8
Or, x3 – 1/x3 = 27/8 + 9/2
Or, x3 – 1/x3 = (27 + 36)/8
Or, 8(x3 – 1/x3) = 63 (Showed)
17. x-1/x = √3 where x≠0
2) Prove that, 23(x2 + 1/x2) = 5(x4 + 1/x4)
Solution:
L.H.S. = 23(x2 + 1/x2)
= 23{(x-1/x)2 + 2.x.1/x}
= 23{(√3)2 + 2}
= 23(3 + 2)
= 23.5
= 115
R.H.S. = 5(x4 + 1/x4)
= 5{(x2)2 + (1/x2)2}
= 5{(x2 + 1/x2)2 – 2.x2.1/x2}
= 5{(5)2 – 2}
= 5.(25 – 2)
= 5.23
= 115
∴ 23(x2 + 1/x2) = 5(x4 + 1/x4) [proved]
3) Find the value of x6 + 1/x6 ?
Solution: From (2), we have,
x2 + 1/x2 = 5
Now,
x6 + 1/x6
= (x2)3 + (1/x2)3
= (x2 +1/x2)3 – 3.x2.1/x2(x2 + 1/x2)
= (5)3 – 3.5
= 125 – 15
= 110 (Ans.)