Question

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Answer 1

Answer:

1.3x+2y=12 , xy=6

(3x+2y)^2 =12^2

=9x^2+4y^2+12xy=144

ie, 9x^2+4y^2+(12×6)=144

=9x^2+4y^2+72=144

9x^2+4y^2=144-72

=72.

2.(3a + 46)^3

Using (x + y)^3 = x^3 + y^3+ 3xy(x+ y)

Where x = 3a & y = 4b

= (3a)^3 + (46)^3 + 3(3a)(4b)(3a + 4b)

= 27a^3 + 646^3 + 3(3a)(4b)(3a) + 3(3a)(4b) (4b)

= 27a^3 + 646^3 + 108a^2b + 144ab

3.x - 1/x = 4 ---------------1)

using this formule , (a -b )² = a² + b² - 2ab

(x - 1/x )² = ( x²+ 1/x ² - 2x×1/x

4² = x² + 1/x² - 2

16+ 2 = x² + 1/x²

18 = x² + 1/x² -------2)

4.We now that,

(a + b + c)2 = a^2+b^2 + c^2+ 2ab+ 2be+ 2ca

15^2 = 83+ 2(ab + be + ca)

225 - 83 = 2(ab + be + ca)

142 = 2(ab + be + ca)

+ ab + bc + ca = 71

Also, we know that,

a^3 + b^3 + c^3 - 3abe = (a + b+ c)(a^2+6^2+ c^2 -ab-be-ca)

= (a+b+ c)(a^2+6^2 + c^2 -(ab+bc+ ca))

= (15)(83 - 71)

= 15 x 12

= 180

5.(2a-b+1)(18a-9b-13)

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