Factorise: (a) 3x2 + 27y2 + z2 - 18xy + 6 √3yz -2 √3zx (b) 27 x3 + 125y3 (c) (2a – 3b + c )2 (d) 1 a3 + b3 + 125 c3 - 15 a b c (e) [x – 1/x y]3 (f) x4 y4 – x y

64^4

(g) 8x3 – (2x – y)3 (h) a6 _ b6

Answer please

step-by-step explanation:

step 1: solve one of the equations for one of the variables. let's solve the first equation for y:

step 2: substitute that equation into the other equation, and solve for x.

step 3: substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.

i think the answer is d im not sure

step by step resolutions

1) n+n+1+n+2=51

3n+3=51

3n=48

n=16

the three numbers are 16, 17 and 18.

2) x+15=4x

3x=15

x=5

radha is now 5 years old

3) the two digits are x and y.

y=3x

(10x+y)+(10y+x)=88

y=3x

11x+11y=88

y=3x

x+y=8

x+3x=8

x=2

the original number can be either 28 or 82.

4) n+(n+2)+(n+4)=48

3n+6=48

3n=42

n=14

the numbers are 14, 16 and 18.

5) y=x+3

10y+x+10x+y=143

y=x+3

11x+11y=143

y=x+3

x+y=13

x+x+3=13

2x=10

x=5

the original number can be 57 or 75.

hope this !