Question

(a).=(x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2​

Answer 1

Step-by-step explanation:

Given

(x²+y²+z²)² - (x²-y²+z²)²

considered

(x²+y²+z²) as = a

and (x²-y²+z²) as = b

now

according to above a²-b² = (a+b)(a-b)

so

(x²+y²+z²)² - (x²-y²+z²)²

= (x²+y²+z²+x²-y²+z²)(x²+y²+z²-x²+y²-z²)

= (2x²+2z²)(2y²)

= 4x²y²+4z²y²

Answer 2

Step-by-step explanation:

(x^2+y^2+z^2)^2-(x^2-y^2+z^2)^2

={x⁴+y⁴+z⁴+2x²y²+2y²z²+2z²x²} - {x⁴+y⁴+z⁴-2x²y²-2y²z²+2z²x²}

=x⁴+y⁴+z⁴+2x²y²+2y²z²+2z²x²- x⁴- y⁴-z⁴+2x²y²+2y²z²-2z²x²

=4x²y²+4y²z² (Ans)

Note: Using rule

• (a+b+c)²= a²+b²+c²+2ab+2bc+2ca
• (a-b+c)²= a²+b²+c²-2ab- 2bc+2ca