Question

(a^2+b^2+c^2)(a^3)+(a^3-b^3-c^3)(b^3)
verify your results for a=1,b=1,c=2
answer fast please​

Answer 1

(a²+b²+c²)(a³)+(a³-b³-c³)(b³)

= (1²+1²+2²)(1³)+(1³-1³-2³)(1³)

= (1+1+4)×1+(1-1-4)×1

= 6×1+(-4)×1

= 6-4

= 2

Answer 2

Question

Find the value of (a²+b²+c²)(a³)+(a³-b³-c³)(b³) where, a = 1, b = 1 and c = 2.

$\rule{300}{0.5}$

Answer

Here we are given the values of 'a', 'b' and 'c'. So a = 1, b = 1 and c = 2.

So first let's substitute the values.

$\sf (a^{2}+b^{2}+c^{2})(a^{3})+(a^{3}-b^{3}-c^{3})(b^{3})$

$\sf (1^{2}+1^{2}+2^{2})(1^{3})+(1^{3}-1^{3}-2^{3})(1^{3})$

Now we have to give the proper value of the numbers. That means that we need to remove the exponents and give it the simplest value.

$\sf (1^{2}+1^{2}+2^{2})(1^{3})+(1^{3}-1^{3}-2^{3})(1^{3})$

$\sf (1+1+4)(1)+(1-1-8)(1)$

Let's solve the brackets now.

$\sf (1+1+4)(1)+(1-1-8)(1)$

$\sf (6)(1)+(-8)(1)$

Now we will remove the brackets and insert multiplication sign where ever required.

$\sf (6)(1)+(-8)(1)$

$\sf 6\times 1+-8\times 1$

Now we have to apply this integer law ⇒ $\sf (+)(-)=(-)$

$\sf 6\times 1+-8\times 1$

$\sf 6\times 1-8\times 1$

Let's do the multiplication part now.

$\sf 6\times 1-8\times 1$

$\sf 6-8$

At last, let's subtract.

$\sf 6-8$

$\sf -2$

∴ (a² + b² + c²)(a³)+(a³ - b³ - c³)(b³) = (-2) when a = 1, b = 1 and c = 1.

$\rule{300}{0.5}$