Factorise the following using appropriate identities:
(i) 9x 2 + 6xy + y 2
(ii) 4y 2 – 4y + 1
(iii) x 2 – \(\frac{y^2 }{ 100}\)
(i) 9x2 + 6xy + y2 = (3x)2 + 2(3x)(y)+(y)2
= (3x + y)(3x + y) [x2 + 2xy + y2 = (x + y)2]
(ii) 4y2 - 4y + 1 = (2y)2 - 2(2y)91) + (1)2
= (2y - 1)(2y - 1) [x2 - 2xy + y2 = (x - y)2]
(iii) x2 -\(\frac{y^2 }{ 100}\) = x2 - (\(\frac{y }{ 10}\))2
= (x + \(\frac{y }{ 10}\)) (x -\(\frac{y }{ 10}\)) [x2 - y2 = (x + y) (x - y)]
If \( x = \left( 2 + \sqrt{3} \right)^3 + \left( 2 - \sqrt{3} \right)^{-3} \) and \( x^3 - 3x + k = 0 \), then the value of \( k \) is:
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD

ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.14). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) ∆ABC ≅ ∠∆BAD
(iv) diagonal AC = diagonal BD [Hint : Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)