Question

if a+b=10,ab=21.then a cube+b cube=

Answer 1

the general formula is
a^3+-b^3=(a+-b)(a^2-+ab+b^2)

in our case will be
a^3+b^3=(a+b)(a^2-ab+b^2)

till now we know only a+b=10 and ab=21

So we have
10(a^2+b^2-21)=

We need to find out a^2+b^2

but a+b=10 /()^2=
a^2+2ab+b^2=100
a^2+2*21+b^2=100
a^2+b^2=100-42
a^2+b^2=58

so
10(58-21)=
10*37=370

Answer 2

a+b=10
ab=21
a=21/b
21/b+b=10
$(21+b ^{2})/b$=10
[tex]21+b ^{2} =10b
 b ^{2} -10b+21=0
 b ^{2} -3b-7b+21=0
 b(b-3)-7(b-3)=0
 b=3,7
 substitute b=3,7 in ab=21
  3a=21
 a=7
 when b=3 then a=7
 7a=21
 a=3
 when b=7 then a=3[/tex]
so,$a^{3} +b^{3}   =7^{3} +3^{3} or 3^{3}+7^{3}   =343+27 or 27 +343   =370$