x³+y³ = (x+y)³−3xy(x+y)

⇒ 12 = 4³−3xy×4

⇒ 12 = 64-12xy

⇒ 12xy = 52

⇒ 3xy = 13 ---- (i)

Also, (x+y)² = x²+y²+2xy

4² = x²+y²+2× (From I)

⇒ x²+y² = 16−=

Therefore, x⁴+y⁴ = (x²+y²)² – 2x²y²

=

=

Hence, option B is the correct answer.

We know that x³ + y³ + z³ − 3xyz = (x + y + z)[(x + y + z)² – 3(xy + yz + zx)]

⇒ 28 – 3×6 = 10[100 – 3(xy + yz + zx)]

⇒ 10/10 = 100 – 3(xy + yz + zx)

⇒ 3(xy + yz + zx) = 100 – 1

⇒ xy + yz + zx = 99/3 = 33

a + 2b = 10 and 2 ab = 9,

(a – 2b)² = (a + 2b)² – 8ab

⇒ (a – 2b)² = 10² – 4 × 9

⇒ (a – 2b)² = 100 − 36

⇒ (a – 2b)² = 64

⇒ (a – 2b) = 8.

3x² – 5x + 1 = 0

⇒ 3x² + 1 = 5x

On dividing by 3x:

⇒ x +  =

On squaring both sides:

⇒ x² +  + 2 × (x) ×  =

⇒ x² +  =  –

⇒ x² +  =  =  =

a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)

⇒ a³ + b³ + c³ – 3abc = (a + b + c)[a(a–b) + b(b–c) + c(c–a)]

Given, a = 43, b = 42, c = 41

⇒ a³ + b³ + c³ – 3abc = (43 + 42 + 42)[43×1 + 42×1 − 41×2]

⇒ a³ + b³ + c³ – 3abc = 378

Given:

⇒

⇒

⇒

Consider [(a² – b²)³ + (b² – c²)³ + (c² – a²)³] ÷ [(a – b)³ + (b – c)³ + (c – a)³]

We know that

If x + y + z = 0

Then x³ + y³ + z³ = 3xyz

Clearly, (a² – b²) + (b² – c²) + (c² – a²) = 0

Hence, [(a² – b²)³ + (b² – c²)³ + (c² – a²)³] = 3(a² – b²)(b² – c²)(c² – a²)

Clearly, (a – b) + (b – c) + (c – a) = 0

Hence, [(a – b)³ + (b – c)³ + (c – a)³] = 3(a – b)(b – c)(c – a)

Now, [(a² – b²)³ + (b² – c²)³ + (c² – a²)³] ÷ [(a – b)³ + (b – c)³ + (c – a)³]

=  = (a + b)(b + c)(c + a)

x² −(√7)x + 1 = 0
x² + 1 = (√7)x

Dividing the equation give in question by x:
x + 1/x = √7

Cubing on both sides:

x³+ 1/x³ + 3(x+1/x) = 7√7

x³ + 1/x³+ 3√7 = 7√7

Hence
x³ + 1/x³= 4√7