Question

The remainder when a¹⁸-b¹⁸ is divisible by a²+ab+b²

use remainder theorem formula pls

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Answer 1

Answer:

The remainder is zero.

To find:

• The remainder when a¹⁸ - b¹⁸ is divisible by a² + ab + b²

Formulas:

• a³ - b³ = (a - b) (a² + ab + b²)

• a² - b² = (a - b) (a + b)

Solution:

Let p (a) = a¹⁸ - b¹⁸

=> p (a) = (a³)⁶ - (b³)⁶

Substitute a³ by n and b³ by m

=> p (a) = n⁶ - b⁶

=> p (a) = (n²)³ - (m²)³

Substitute n² by p and m² by q

=> p (a) = p³ - q³

=> p (a) = (p - q) (p² + pq + q²)

Resubstituting p = n² and q = m²

=> p (a) = (n² - m²) (n⁴ + n²m² + m⁴)

=> p (a) = (n - m) (n + m) (n⁴ + n²m² + m⁴)

Resubstituting n = a³ and m = b³

=> p (a) = (a³ - b³) (a³ + b³) (a¹² + a⁶b⁶ + b¹²)

=> p (a) = (a - b) (a² + ab + b²) (a³ + b³)

(a¹² + a⁶b⁶ + b¹²)

While dividing, p (a) by a² + ab + b² we get, a² + ab + b² is a factor of a¹⁸ - b¹⁸

Hence, the remainder will be zero.