Question

A+b+c=5 nad ab+bc+ca=10 , then find the value of a3+b3+c3-3abc

Answer 1

(a+b+c)²=a²+b²+c²+2(ab+bc+ca)

(5)²=a²+b²+c²+2(10)

25=a²+b²+c²+20

a²+b²+c²=5

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

=(5)(5-10)

=(5)(-5)

=-25

Answer 2

Answer:

The correct answer is $&a^{3}+b^{3}+c^{3}-3 a b c=-25$.

Step-by-step explanation:

A binomial expansion exists as a technique used to allow us to expand and simplify algebraic expressions in the form into a totality of terms of the form.

The binomial theorem specifies the expansion of any power $(a + b)^{m}$ of a binomial $(a + b)$ as a certain sum of products $a^{i} b^{j}$, such as $(a + b)^{2} = a^{2} + 2ab + b^{2}$.

Step 1

We have,

$&a+b+c=5$

$&a b+b c+c a=10$

Since,

$&(a+b+c)^{2}=25$

$&a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=25$

Substituting the values in the above equation,

$&a^{2}+b^{2}+c^{2}+2 \times 10=25$

$&a^{2}+b^{2}+c^{2}=5$

Step 2

We know that,

$&a^{3}+b^{3}+c^{3}-3 a b c=(a+b+c)\left(a^{2}+b^{2}+c^{2}-(a b+b c+c a)\right)$

Substituting the values in the above equation,

$&a^{3}+b^{3}+c^{3}-3 a b c=(5) \times(5-10)$

$&a^{3}+b^{3}+c^{3}-3 a b c=5 \times-5$

$&a^{3}+b^{3}+c^{3}-3 a b c=-25$

Therefore, the correct answer is $&a^{3}+b^{3}+c^{3}-3 a b c=-25$.

#SPJ2