Question

a2+b2=69 ,a4+b4=3409 then a3+b3=?​

Answer 1

Answer:

Step-by-step explanation:

given,

a^2+b^2=69,

(a+b)^2-2ab=69----(1),

a^4+b^4=3409,

(a^2+b^2)^2-2a^2b^2=3409,

69)^2-2a^2b^2=3409(substitute the value of a^2+b^2=69 ),

4761-2a^2b^2=3409,

2a^2b^2=4761-3409,

2a^2b^2=1352,

a^2b^2=1352/2,

(ab)^2=676,

ab=26----(2),

substitute (2) in (1),

(a+b)^2-2*26=69,

(a+b)^2=69+52,

a+b^2=121,

a+b=11,

a^3+b^3=(a+b)(a^2+b^2-ab),

=(11)(69-26),

=11*43,

=473,

Hope it helps