Question

find (x-y)³ +(y-z)³ + (z-x)³

find x+1/x if x² +1/x²=34

factorise x²-y²x-y

find the value of x³+y³+12xy - 64 when x+y=4

if x and y are two positive real numbers such that x² - 4y²=40 xy=6 find x+2y

if abc are all nonzero and a+b+c=0 prove that a²/bc+b²/ac+c²/ab=3

(m+2n)²+101(m+2n)+100

Answer 1

1).When a+b+c = 0 we know that a³ + b³ + c³ = 3 a b c

(x-y)³ + (y-z)³ + (z - x)³
= 3 (x - y) (y -z) (z- x)
= 3 (xy - x z -y² + yz) (z -x)
= 3 (xyz - x² y - x z² + x² z - y² z + x y² + y z² - xyz)
= 3 [ x² (z - y) + z² (y - x) + y² (x - z) ]

2) x^2+(1/x^2)=34
(x+1/x)^2-2=34
(x+1/x)^2=36
(x+1/x)=6 or -6
3) I think the question went wrong the question may be

x 2 − y 2 −x+y= (x 2 − y 2)- (x−y) =(x−y)(x+y)−(x−y) =(x−y)(x+y−1) x2−y2−x+y=(x2−y2)−(x−y)=(x−y)(x+y)−(x−y)=(x−y)(x+y−1) 4)Cubing both sides of the equation (x+y)^3=x^3+y^3+ 3xy(x+y)=4^3=64 Substituting for x+y as 4 on the left hand side x^3+y^3+3xy4=64 x^3+y3+12xy=64 So x^3+y^3+12xy-64=0 5) x² + 4y² = (x + 2y)² - 4xy x² + 4y² = 40 ------> (x + 2y)² - 4xy =40 (x + 2y)² - 24 = 40 (x + 2y)² = 64 ----> x + 2y = +8 or -8 6)Given a + b + c = 0 ⇒ a3 + b3 + c3 = 3abc → (1) Consider, (a2/bc) + (b2/ca) + (c2/ab) = (a3 + b3 + c3)/abc = 3abc/abc = 3 [From (1)].. 7)Put m+2n=x then we get equation is x^2+101x+100=0 x^2+100x+x+100=0 x(x+100)+1(x+100)=0 (x+100)(x+1)=0 now put the value of x in above equation (m+2n+100)(m+2n+1)=0 ​​​​​