Mathematics
Assertion (A): (26)3 + (−15)3 + (−11)3 = 3 × 26 × 15 × 11.
Reason (R): If x + y + z = 0, then x3 + y3 + z3 = 3xyz
A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false
Answer
We know that,
⇒ x3 + y3 + z3 - 3xyz = (x + y + z)(x2 + y2 + z2 - xy - yz - zx)
If x + y + z = 0, then :
⇒ x3 + y3 + z3 - 3xyz = 0
⇒ x3 + y3 + z3 = 3xyz.
So, reason (R) is true.
⇒ 26 + (-15) + (-11)
⇒ 26 - 26
⇒ 0
Since, 26 + (-15) + (-11) = 0,
∴ (26)3 + (−15)3 + (−11)3 = 3 × 26 × -15 × -11
Assertion (A) is false.
Thus, A is false and R is true.
Hence, Option 2 is the correct option.
Related Questions
If l + m − n = 9 and l2 + m2 + n2 = 31, then mn + nl − lm is :
-25
25
-2
-5
If a − b + c = 6 and a2 + b2 + c2 = 38, then ab + bc − ca is :
0
1
-1
not possible
Assertion (A): (1 - 3x)3 can be expanded as 1 - 27x3 - 9x - 27x2.
Reason (R): (a - b)3 = a3 - b3 - 3ab(a - b)A is true, R is false
A is false, R is true
Both A and R are true
Both A and R are false