Question

Please help me out ​

Answer 1

Solution

Given :-

• a + b + c = 8 ---------(1)
• ab + bc + ca = 11 ------(2)

Find :-

• Value of ( a³ + b³ + c³ - 3abc)

Explanation

Important Formula

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

Now, Squaring both Side of equation(1)

==> (a + b + c)² = 8²

==> ( a²+b²+c²+2ab+2bc+2ca) = 64

==> (a²+b²+c²)+2(ab+bc+ca) = 64

keep value by equ(2)

==> (a²+b²+c²)+2(11) = 64

==> (a²+b²+c²) = 64 - 22

==> (a²+b²+c²) = 42 ----------(3)

So, Now Calculate ( a³ + b³ + c³ - 3abc)

Keep Value by equ(1),(2) & (3)

==> ( a³ + b³ + c³ - 3abc) = (8) * (42 - 11)

==> ( a³ + b³ + c³ - 3abc) = 8 * 31

==> ( a³ + b³ + c³ - 3abc) = 248

Hence

• Value of ( a³ + b³ + c³ - 3abc) will be 248

_________________

Answer 2

Given ,

a + b + c = 8 and

ab + bc + ac = 11

We need to find the value of a³ + b³ + c³ - 3abc

Here we need to use a formula , i.e.,

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

( a + b + c ) = 8

Squaring on both sides , we get ,

⇒ ( a + b + c )² = 64

⇒ a² + b² + c² + 2ab + 2bc + 2ac = 64

⇒ ( a² + b² + c² ) = 64 - 2 ( ab + bc + ac )

⇒ ( a² + b² + c² ) = 64 - 22

⇒ ( a² + b² + c² ) = 42 ... (1)

Now our required ,

$\implies \bold{(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)}\\\\\implies \bold{(a^3+b^3+c^3-3abc)=8(42-11)\;[\;From\;(1)\;]}\\\\\implies \bold{(a^3+b^3+c^3-3abc)=8*31}\\\\\implies \bold{(a^3+b^3+c^3-3abc)=248}$