1.

a^6 %2B 32*a^3 - 64

Answer»

a^6+ 32a³– 64

a^6+ 8a³– 64 + 24a³

⇒(a²)³+ (2a)³+ (- 4)³- 3×(a²)×(2a)×(- 4) … (i)

We use the identity

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

substitution in (i) we get

(a²)³+ (2a)³+ (- 4)³- 3×(a²)×(2a)×(- 4) = (a²+ 2a - 4) (a⁴+ 4a²+ 16 - 2a³+ 8a + 4a²)

⇒(a²+ 2a - 4) (a⁴+ 8a²+ 16 - 2a³+ 8a)

(a²+ 2a - 4) (a⁴- 2a³+ 8a²+ 8a + 16)



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