Question

Which of the equations are true identities? A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n B. ( x + 1 ) 2 − 2 x + y 2 = x 2 + y 2 + 1

Answer 1

Both A and B are true identities

Explanation:

A. N ( n − 2 ) ( n + 2 ) = n 3 − 4 n

We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)

So,

n ( n − 2 ) ( n + 2 ) = n(n² - 2²) (difference of two squares)

= n³ - 2²n (expanding the brackets)

= n³ - 4n (simplifying)

So, L.H.S = R.H.S

B. ( x + 1 )² − 2x + y² = x² + y² + 1

We need to show that (left-hand-side)L.H.S = R.H.S (right-hand-side)

So,

( x + 1 )² − 2x + y² = x² + 2x + 1 - 2x + y² (expanding the brackets)

= x² + 2x - 2x + 1 + y² (collecting like terms)

= x² + 1 + y²

= x² + y² + 1 (re-arranging)

So, L.H.S = R.H.S

So, both A and B are true identities since we have been able to show that L.H.S = R.H.S in both situations.