Question

8)establish the formula ( a+b+c)² = a² + b² + c2²+ 2( ab +bc + ca ) and give 4
examples​

Answer 1

$\underline{\textbf{\large{Step-by-step explanation:}}}$

it Can be establish as _⬇

( a + b + c) ²

= (a + b + c) (a + b + c)

=a² + ab + ac + ab + b² +bc + ac + ab + c²

= a² + b² + c² + ab + ac + ab + bc + ac + bc + ab

= a² + b² + c² + ab + ab + ac + ac + bc + ac

= a² + b² + c² + 2ab + 2ac + 2bc

= a² + b² + c² + 2 (ab + ac + bc)

hence the required formula is

( a + b + c) ² = a² + b² + c² + 2 (ab + bc + ac )

$\underline{\textbf{\large{Examples :}}}$

1)) ( x + y + z) ²

.= ( x² + y² + z² + 2(xy + yz + xz))

2)) (2z + 0 + x)²

= ( (2z)² + (0)² + (x)² + 2((2z)(0) + (0)(x) + (2z) (x))

= 4z² + 0 + x² + 0 + 0 + 4zx

= 4z² + x² + 4zx

= 4z² + x² + 4zx

3)) ( 2 + x + √3 )²

= (2)² + (x)² + (√3)² + 2(2x + x√3 + 2√3 )

= 4 + x² + 3 + 4x + 2√3x + 4√3

= 7 + x² + 2(2x+ √3x) +4√3

= x² + 2(2x + √3x) + 7 + 4√3

4)) ( 2x + (1/2) - z) ²

= (2x + (1/2) +(-z)) ²

= ((2x)² + (1/2)² +(-z) ² + 2 ((2x) (1/2)+(1/2)(-z) +(2x) (-z))

= 4x² + (1/4)+z² + 2x +(-z) +4(x )(-z)

=4x² + (1/4) + z² + 2x -z - 4zx

=4x² + 2x +z² - 4zx -z + (1/4)

= 4x² + 2x +z² - 4zx -z + (1/4)

= 4x² + 2x +z² - 4zx -z + (1/4)

= 4x² + 2x + z² -z (4x + z) +(1/4)