Question

Please guys solve--------

Topic :-Polynomial.....


I want right answer........

Answer 1

a+b=8 and a²+b² = 40
then ,
(a+b)²=a²+b²+2ab
8²=40+2ab
24 = 2ab
ab = 24/2 = 12
now , a³+b³ = (a+b)(a²+b²-ab)
8(40-12) = 8(28) = 224




P=4-q
p³+q³+12pq=64.
P+q-4= 0
p³+q³-64+12pq = 0
using formula a³ + b³ + c³-3abc = (a+b+c)(a²+b²+ c²-ab-bc-ca)
(P)³+(q)³+(-4)³-3(P)(q)(-4) = [P+q+(-4)](p²+q²+16-pq+4q+4p)
but P+q-4 = 0
so ,
p³+q³-64+12pq = 0
next question will be solved by this method only if any problem , tell me in messages or comments section
hope this helps

Answer 2

p=4-q

p+q=4

Cubing both the sides

(p+q)^3=4^3

p^3+q^3+3pq(p+q)=64

p^3+q^3+3pq(4)=64

p^3+q^3+12pq=64 Hence Proved