To find the product of three binomials, we can use the formula given below.

(x+a) (x+b) (x+c) = x^{3}+(a+b+c)x^{2} + (ab+bc+ca)x + abc

Find the expansion of

**Problem 1 :**

(x+1)(x+4)(x+7)

**Solution :**

(x+a)(x+b)(x+c) = x^{3}+(a+b+c)x^{2} + (ab+bc+ca)x + abc

a = 1, b = 4 and c = 7

= x^{3}+ (1+4+7)x^{2}+[1(4)+4(7)+7(1)]x+1(4)(7)

= x^{3} + 12x^{2} + (4+28+7)x + 28

= x^{3} + 12x^{2} + 39x + 28

**Problem 2 :**

(p+2)(p-4)(p+6)

**Solution :**

x = p, a = 2, b = -4 and c = 6

p^{3} + (2-4+6)p^{2} + [2(-4)+(-4)6+6(2)]p + (2)(-4)(6)

= p^{3} + 4p^{2} + (-8-24+12)p - 48

= p^{3} + 4p^{2} - 20p - 48

**Problem 3 :**

(x+5)(x-3)(x-1)

**Solution :**

Here x = x, a = 5, b = -3 and c = -1

= x^{3} + (5-3-1)x^{2} + [5(-3)+(-3)(-1)(-1)5]x + 5(-3)(-1)

= x^{3} + x^{2} + (-15+3-5)x + 15

= x^{3} + x^{2} - 17x + 15

**Problem 4 :**

(x-a)(x-2a)(x-4a)

**Solution :**

Here x = 3x, a = -a, b = -2a and c = -4a

= x^{3} + (-a-2a-3a)x^{2} +

[(-a)(-2a)+(-2a)(-3a)+(-3a)(-a)]x + (-a)(-2a)(-3a)

= x^{3}-6ax^{2} + (2a^{2}+6a^{2}+3a^{2})x-6a^{3}

= x^{3}-6ax^{2} + 11a^{2}x - 6a^{3}

**Problem 5 :**

(3x+1)(3x+2)((3x+5)

**Solution :**

Here x = 3x, a = 1, b = 2 and c = 5

= (3x)^{3} + (1+2+5)(3x)^{2} + (1.2+2.5+5.1) (3x) + 1.2.5

= 27 x^{3} + (8)9x^{2} + (2+10+5)(3x)+1.2.5

= 27 x^{3}+ 72x^{2} + 51x+ 10

**Problem 6 :**

(2x+3)(2x-5)(2x-7)

**Solution :**

Here x = 2x, a = 3, b = -5 and c = -7

= (2x)^{3} + (3-5-7)(2x)^{2} + [(3)(-5)+(-5)(-7)+(-7)(3)] (2x) + (3)(-5)(-7)

= 8x^{3} + (-9)(4x^{2}) + (-15+35-21)(2x) + 105

= 8x^{3} - 36 x^{2} - 2x + 105

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