Question

- यदि t^2 - 4t +1= 0, तो t^3+1/t^3का मान क्या होगा?
13​

Answer 1

Answer:

Step-by-step explanation:

$t^2-4t +1=0\\\\t^2+1=4t\\\\\frac{t^2+1}{t} =\frac{4t}{t}\\\\t+\frac1t=4 ---------(1)$

दोनों ओर  क्यूब लेने पर

$(t+\frac1t)^3 =4^3\\\\t^3+\frac{1}{t^3}+3(t)(\frac1t)(t+\frac1t)=64\\\\t^3+\frac{1}{t^3}+3(4) =64 \,\,\, ( \because t+\frac1t=4)\\\\t^3+\frac{1}{t^3} =64-12=52$

अतः

$t^3 + \frac {1}{t^3} =52$