**Samacheer Kalvi 10th Maths Book Back Solution:**

Tamil Nadu **10th Maths Book Back Answers Unit 3 – Algebra Ex 3.3** are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete** Samacheer Kalvi Books Back Answers/Solutions** are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 3 Algebra consists of 19 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Algebra Ex 3.3 Answers in English.

**10th Maths Book Back Answers/Solutions:**

TN Samacheer Kalvi 10th Maths Unit 3 Chapter 3 Book Back Exercise has given below. The 10th Maths Book Back Solutions Guide is uploaded below,

**Chapter 3**

**Exercise 3.3 Algebra**

1.Find the LCM and GCD for the following and verify that f(x) × g(x) = LCM × GCD.

(i) 21x^{2}y, 35xy^{2}

(ii) (x^{3} – 1)(x + 1), x^{3} + 1

(ii) (x^{3} – 1) (x + 1), (x^{3} – 1)

(iii) (x^{2}y + xy^{2}), (x^{2} + xy)

**Solution:
**(i) f(x) = 21x

^{2}y = 3 × 7x

^{2}y

g(x) = 35xy

^{2}= 7 × 5xy

^{2}

G.C.D. = 7xy

L.C.M. = 7 × 3 × 5 × x

^{2}y

^{2}= 105x

^{2}× y

^{2}

L.C.M × G.C.D = f(x) × g(x)

105x

^{2}y

^{2}× 7xy = 21x

^{2}y × 35xy

^{2}

735x

^{3}y

^{3}= 735x

^{3}y

^{3}

Hence verified.

(ii) (x^{3} – 1)(x + 1) = (x – 1)(x^{2} + x + 1)(x + 1)

x^{3} + 1 = (x + 1) (x^{2} – x + 1)

G.C.D = (x+ 1)

L.C.M = (x – 1)(x + 1)(x^{2} + x + 1)(x^{2} – x + 1)

∴ L.C.M. × G.C.D = f(x) × g(x)

(x – 1)(x + 1)(x^{2} + x + 1) (x^{2} – x + 1) = (x – 1)

(x^{2} + x + 1) × (x + 1) (x^{2} – x + 1)

(x^{3} – 1)(x + 1)(x^{3} + 1) = (x^{3} – 1)(x + 1)(x^{3} + 1)

∴ Hence verified.

(iii) f(x) = x^{2}y + xy^{2} = xy(x + y)

g(x) = x^{2} + xy = x(x + y)

L.C.M. = x y (x + y)

G.C.D. = x (x + y)

To verify:

L.C.M. × G.C.D. = xy(x + y) × (x + y)

= x^{2}y (x + y)^{2} ……….. (1)

f(x) × g (x) = (x^{2}y + xy^{2})(x^{2} + xy)

= x^{2}y (x + y)^{2} …………… (2)

∴ L.C.M. × G.C.D = f(x) × g{x).

Hence verified.

2.Find the LCM of each pair of the following polynomials

(i) a^{2} + 4a – 12, a^{2} – 5a + 6 whose GCD is a – 2

(ii) x^{4} – 27a^{3}x, (x – 3a)^{2} whose GCD is (x – 3a)

**Solution:**

(i) f(x) = a^{2} + 4a – 12 = (a + 6)(a – 2)

(ii) f(x) = x^{4} – 27a^{3}x = x(x^{3} – (3a)^{3})

g(x) = (x – 3a)^{2}

G.C.D = (x – 3a)

L.C.M. × G.C.D = f(x) × g(x)

L. C.M = x(x3−(3a)3)×(x−3a)2(x−3a)

L.C.M = x(x^{3} – (3a)^{3}) . (x – 3a)

= x(x – 3a)^{2} (x^{2} + 3ax + 9a^{2})

3.Find the GCD of each pair of the following polynomials

(i) 12(x^{4} – x^{3}), 8(x^{4} – 3x^{3} + 2x^{2}) whose LCM is 24x^{3} (x – 1)(x – 2)

(ii) (x^{3} + y^{3}), (x^{4} + x^{2}y^{2} + y^{4}) whose LCM is (x^{3} + y^{3}) (x^{2} + xy + y^{2})

**Solution:**

(i) f(x)= 12(x^{4} – x^{3})

g(x) = 8(x^{4} – 3x^{3} + 2x^{2})

L.C.M = 24x^{3} (x – 1)(x – 2)

(ii) (x^{3} + y^{3}), (x^{4} + x^{2}y^{2} + y^{4})

L.C.M. = (x^{3} + y^{3})(x^{2} + xy + y^{2})

4.Given the LCM and GCD of the two polynomials p(x) and q(x) find the unknown polynomial in the following table

**Solution:**

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