Question

factorise a^4-a^13plz ans this​

Answer 1

Answer:

$a^{4}-a^{13}\\=a^{4}(1-a)(1+ a+a^{2})(1+a^{3}+a^{6})$

Step-by-step explanation:

$a^{4}-a^{13}\\=a^{4}(1-a^{9})\\=a^{4}[1-(a^{3})^{3})]\\=a^{4}[(1-a^{3})[1^{2}+1\times a^{3}+(a^{3})^{2}]$

$=a^{4}(1-a)(1+1\times a+a^{2})(1+a^{3}+a^{6})$

$=a^{4}(1-a)(1+ a+a^{2})(1+a^{3}+a^{6})$

Therefore,

$a^{4}-a^{13}\\=a^{4}(1-a)(1+ a+a^{2})(1+a^{3}+a^{6})$

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Answer 2

a⁴ (1 - a) ( 1 + a + a² ) ( 1 + a³ + a⁶ )

Step-by-step explanation:

Given,

a⁴ - a¹³

Taking a⁴ as common:

a⁴ (1 - a⁹)

a⁴ ( (1)³ - (a³)³ )

Substituting the formula: a³ - b³ = (a - b) (a² + ab + b²)

a⁴ ( (1 - a³) (1)² + (1 × a³) + (a³)² )

a⁴  (1 - a³) ( 1 + a³ + a⁶ )

a⁴ ( (1)³ - (a)³ ) ( 1 + a³ + a⁶ )

Substituting the formula: a³ - b³ = (a - b) (a² + ab + b²)

a⁴ (1 - a) ( (1)² + (1 × a) + (a)² ) ( 1 + a³ + a⁶ )

a⁴ (1 - a) ( 1 + a + a² ) ( 1 + a³ + a⁶ )

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