Factorise:
(i) 4x 2 + 9y 2 + 16z 2 + 12xy – 24yz – 16xz
(ii) 2x 2 + y 2 + 8z 2 – 2√2 xy + 4√2 yz – 8xz
(i) It is known that,
(x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2zx
4x2 + 9y2 + 16z2 + 12xy – 24yz – 16xz : (2x)2 + (3y)2 + (-4z)2 + 2(2x)(3y) + 2(3y)(-4z) + 2(2x)(-4z)
= (2x + 3y - 4z)2 = (2x + 3y - 4z)(2x + 3y - 4z)
(ii) 2x2 + y2 + 8z2 – 2√2xy + 4√2 yz – 8xz
= (-√2x)2 + (y)2 + (2√2z)2 + 2(-√2x) (y) + 2(y)(2√2z) + 2 (-√2x)(2√2z)
= (-√2x + y + 2√2z)2 = (-√2x + y + 2√2z) (-√2x + y + 2√2z)
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)