Question

factorise a^6-18a^3+125 ​

Answer 1

a^6-18a³+125=0

(a²)^3 + (-3a)^3 + (5)^3 - 3(a²)(-3a)(5) 

by applying,

a^3 + b^3 + c^3 - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca) 

Here a = a²; b = -3a and c = 5 

a^6 + 18a^3 + 125 = (a²)^3 + (-3a)^3 + (5)^3 - 3(a²)(-3a)(5) 

(a² - 3a + 5){(a^4) + 9a² + 25 + 3(a^3) + 15a - 5a²} 

(a² - 3a + 5){(a^4) + 3(a^3) + 4a² + 15a + 25}

no two such factors can be found

Answer 2

Answer:

this can be solved by this formula-

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

Step-by-step explanation:

a⁶-18a³+125

= a⁶+27a³+125-45a³

=(a²)³+(3a)³+(5)³-3×(a²)×(3a)×(5)

=(a²+3a+5)×{(a²)²+(3a)²+(5)²-(a²)×(3a)-(3a)×5-5×a²}

=(a²+3a+5)(a⁴+9a²+25-3a³-15a-5a²)

=(a²+3a+5)(a⁴-3a³+4a²-15a+25)

here your answer is here I hope it was helpful