Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4E Solutions
EXERCISE – 4E
(Q1) Simplify:
(i) (x + 6) (x + 4) (x – 2)
Solution:
{(x + 6) (x + 4)} (x – 2)
= {x (x + 4) + 6 (x + 4)} (x – 2)
= {x2 + 4x + 6x + 24} (x – 2)
= {x2 + 10x + 24} (x – 2)
= x × (x2 + 10x + 24) – 2 (x2 + 10x + 24)
= x3 + 10x2 + 24x – 2x2 – 20x – 48
= x3 + 8x2 + 4x – 48
(ii) (x – 6) (x – 4) (x + 2)
Solution:
(x – 6) (x – 4) (x + 2)
= {(x – 6) (x – 4)} (x + 2)
= {x (x – 4) – 6 (x – 4)} (x + 2)
= {x2 – 4x – 6x + 24} (x + 2)
= x (x2 – 10x + 24) + 2 (x2 – 10x + 24)
= x3 – 10x2 + 24x + 2x2 – 20x + 48
= x3 – 18x2 + 4x + 48
(iii) (x – 6) (x – 4) (x – 2)
Solution:
(x – 6) (x – 4) (x – 2)
= {(x – 6) (x – 4)} (x – 2)
= {x (x – 4) – 6 (x – 4)} (x – 2)
= {x2 – 4x – 6x + 24} (x – 2)
= {x2 – 10x + 24} (x – 2)
= x (x2 – 10x + 24) -2 (x2 – 10x + 24)
= x3 – 10x2 + 24x – 2x2 + 20x – 48
= x3 – 12x2 + 44x – 48
(iv) (x + 6) (x – 4) (x – 2
Solution:
(x + 6) (x – 4) (x – 2)
= {(x + 6) (x – 4)} (x – 2)
= {x (x – 4) + 6 (x – 4)} (x – 2)
= {x2 – 4x + 6x – 24} (x – 2)
= {x2 + 2x – 24} (x – 2)
= x (x2 + 2x – 24) – 2 (x2 + 2x – 24)
= x3 + 2x2 – 24x – 2x2 – 4x + 48
= x3 – x2 – 28x + 48
= x3 – 28x + 48
(Q2) Simply using following identity
(a + b) (a2 ± ab + b2) = a3 ± b3
(i) (2x + 3y) (4x2 – 6xy + 9y2)
Solution:
(2x + 3y) (4x2 – 6xy + 9y2)
∴ a = 2x, b = 3y
= (2x + 3y) ((2x)2 – (2x) (3y) + (3y)2)
= (2x)3 ± (3y)3
= 8x3 ± 27y3
(ii) (3x – 5/x) (9x2 + 15 + 25/x2)
Solution:
(3x – 5/x) (9x2 + 15 + 25/x2)
a = 3x, b = 5/x
= (3x – 5/x) ((3x)2 – (3x) (5/x) + (+5/x)2)
= (3x)3 + (+5/x)3
= 27x3 + 125/x3
(iii) (a/3 – 3b) (a2/9 + ab + 9b2)
Solution:
(a/3 – 3b) (a2/9 + ab + 9b2)
=> a = a/3, b = 3b
=> (a/3 – 3b) ((a/3)2 + (a/3) (3b) + (3b)2)
=> (a/3)3 = (3b)3
=> a3/27 = 27b3
(Q3) Using suitable identity, evaluate
(i) (104)3
Solution:
(104)3
We have to use,
(a + b)3 = a3 + 3ab (a + b) + b3
First we have to convert into (a + b)
(104) = (100 + 4)
∴ (100 + 4)3 = (100)3 + 3 (100) (4) (100 + 4) + (4)3
= 1,000,000 + 1,200 (104) + 64
= 1,000,000 + 124,800 + 64
= 1,124,864
(ii) (97)3
Solution:
(97)3
We have to use,
(a + b)3 = a3 + 3ab (a + b) + b3
First we have convert into (a + b)
(97) = (90+7)
∴ (90+7)3 = (90)3 + 3 (90) (7) (90+7)+ (7)3
= 729,000 + 1890 (97) + 343
= 729,000 + 183,330 + 343
= 912,673
(Q4) Simply:
(x2 – y2)3 + (y2 – z2)3 + (z2 – x2)3/(x – y)3 + (y-z)3 + (z – x)3
Solution:
(x2 – y2)3 + (y2 – z2)3 + (z2 – x2)3/(x – y)3 + (y-z)3 + (z-x)3
We know that, If a + b + c = 0 then a3 + b3 + c3 = 3ab
∴ a = x2 – y2, b = y2 – z2, c = z2 – x2
∴ x2 – y2 + y2 – z2 + z2 – x2 = 0
= 3 (x2-y2) (y2 – z2) (z2 – x2)/3(x – y) (y-z) (z-x)
= (x+y) (x-y) (y-z (y+z) (z-x) (z + x )/(x – y) (y-z) (z – x)
{(a2 – b2) = (a – b) (a + b)}
= (x + y) (y + z) (z + x)
(Q5) Evaluate:
(i) 0.8×0.8×0.8+0.5×0.5×0.5/0.8×0.8-0.8×0.5+0.5×0.5
(ii) 1.2×1.2+1.2×0.3+0.3×0.3/1.2×1.2×1.2-0.3×0.3×0.3
Solution:
(i) 0.8×0.8×0.8+0.5×0.5×0.5/0.8×0.8-0.8×0.5+0.5×0.5
= (0.8)3+(0.5)3/[(0.8)2 – 0.8×0.5+(0.5)2]
= (0.8+0.5) [(0.8)2 – 0.8×0.5 + (0.5)2]/[(0.8)2 – (0.8) × (0.5) + (0.5)2}
∵ ((a3 – b3) = (a + b) (a2 – ab + b2))
= 0.8 + 0.5
= 1.3
(ii) 1.2×1.2+1.2×0.3+0.3×0.3/1.2×1.2-0.3×0.3×0.9
= [(1.2)2+ 1.2×0.3 + (0.3)2]/[(1.2)3 – (0.3)3]
= Now, we have to use,
(a3 – b3) = (a + b) (a2 – ab + b2)
= [(1.2)2 + 1.2×0.3 + (0.3)2]/(1.2÷0.3) ((1.2)2 + (1.2) × (0.3) + (0.3)2]
= 1/1.2-0.3
= 1/0.9
= 1/0.9×10-1
= 10/9
= 1 1/9
(Q6) If a – 2b + 3c = 0; State the value of a3 – 8b3 + 27c3
Solution:
Given that,
a + (-2b) + 3c = 0
∴ a3 – 8b3 + 27c3 = 3×a×(-2b) × (3c)
= -18abc
∵ (a3 + b3 + c3 = 3abc)
(Q7) If x + 5y = 10; find the value of x3 + 125y3 + 150xy – 1000
Solution:
Given that,
x + 5y = 10 — (1)
Cubing on both sides of equation (1)
(x + 5y)3 = (10)3
(x)3 + 3 (x) (5y) (x + 5y) + (5y)3 = 1000
x3 + 15xy (10) + 125y3 = 1000 (∵ from (1)}
x3 + 125y3 + 150xy – 1000 = 0
(Q8) If x = 3 + 2√2, find:
1/x (iii) (x – 1/x)3
x – 1/x (iv) x3 – 1/x3
Solution:
Given that, x = 3 + 2√2
1/x = 1/3+2√2
= 1/3+2√2 × 3-2√2/3-2√2
Rationalising denominator and numerator
= 3 – 2√2/(3+2√2)(3-2√2)
= 3-2√2/(3)2-(2√2)2
{∵ (a + b) (a – b) = a2 – b2}
= 3-2√2/9-(2×2×2)
= 3-2√2/9-8
= 3-2√2/1 = 3-2√2
(ii) x – 1/x
= x – 1/x = (3 + 2√2) – 1/(3+2√2)
= (3 + 2√2) – (3 – 2√2)
= 3 + 2√2 – 3 + 2√2
= 4√2 — (A)
(iii) (x – 1/x)3
= (x – 1/x)3 = (4√2)3 (From A)
= 64 × √2 × √2 × √2
= 64 2√2
= 128 √2
(iv) x3 – 1/x3
= (x – 1/x)3 = x3 – 1/x3 – 3 × x × 1/x (x – 1/x)
128√2 = x3 – 1/x3 – 3 (4√2)
128√2 ÷ = x3 – 1/x3 – 12√2
128√2 + 12√2 = x3 – 1/x3
140√2 = x3 – 1/x3
∴ x3 – 1/x3 = 140√2
(Q9) If a + b = 11 and a2 + b2 = 65; find a3 + b3
Solution:
Given, a + b = 11 and a2 + b2 = 65
Now, we have to use,
(a + b)2 = a2 + 2ab + b2
(11)2 = (a2 + b2) + 2ab
121 = 65 + 2ab
121 – 65 = 2ab
56/2 = ab
ab = 28 — (A)
∴ (a + b)3 = a3 + b3 + 3ab (a + b)
(11)3 = a3 + b3 + 3 (28 (11) (From A)
1331 = a3 + b3 + 33×28
1331 – 924 = a3 + b3
407 = a3 + b3
∴ a3 + b3 = 407
(Q10) Prove that:
x2 + y2 + z2 – xy – yz – zx is always positive
Proof:
x2 + y2 + z2 – (xy + yz + zx)
Let, x = 1, y = 2, z = 3
= (1)2 + (2)2 + (3)2 – (1 (2) + (2) (3) + (3) (1))
= (1 + 4 + 9) – (2 + 6 + 3)
= 14 – 11
= 3
∴ x2 + y2 + z2 > xy + yz + zx
Therefore, it will always be positive.
(Q11) Find:
(i) (a + b) (a + b)
Solution:
(a + b) (a + b)
= a (a+ b) + b (a + b)
= a2 + ab + ba + b2
= a2 + 2ab + b2
(ii) (a + b) (a + b) (a + b)
Solution:
(a + b) (a + b) (a + b)
= [(a + b) (a + b)] (a + b)
= [a2 + ab + ba + b2] (a + b)
= (a2 + 2ab + b2) (a + b)
= a×(a2 + 2ab + b2) + b × (a2 + 2ab + b2)
= a3 + 2a2b + ab2 + ba2 + 2ab2 + b3
= a3 + b3 + 3a2 b + 3ab2
(iii) (a – b) (a – b) (a – b) by using the result of part (ii)
Solution:
(a – b) (a – b) (a – b)
= a3 + 3a2 (-b) + 3a (-b)2 + (-b)3
= a3 – 3a2b + 3ab2 – b3
Here is your solution of Selina Concise Class 9 Maths Chapter 4 Expansions Exercise 4E
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