Check sibling questions

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Misc 1 Evaluate: (𝑖^18+(1/i)^25 )^3 (𝑖^18+(1/𝑖)^25 )^3 = (𝑖^18+ 1/(π’Š)^πŸπŸ“ )^3 = (𝑖^18+ 1/(π’Š Γ— π’Š^πŸπŸ’ ))^3 = ((π’Š^𝟐 )^πŸ—+1/(𝑖 Γ— (π’Š^𝟐 )^𝟏𝟐 ))^3 Putting i2 = βˆ’πŸ = ((βˆ’πŸ)^πŸ—+1/〖𝑖 Γ— (βˆ’πŸ)γ€—^12 )^3 = (βˆ’πŸ+1/(𝑖 Γ— 𝟏))^3 = (βˆ’1+1/𝑖)^3 Removing π’Š from the denominator = (βˆ’1+1/π‘–Γ—π’Š/π’Š)^3 = (βˆ’1+𝑖/π’Š^𝟐 )^3 = (βˆ’1+𝑖/((βˆ’πŸ)))^3 = (βˆ’πŸ – π’Š )πŸ‘ = (βˆ’1(1+ 𝑖 ))3 = (βˆ’πŸ)πŸ‘ (𝟏 + π’Š )πŸ‘ = (βˆ’1)(1 + 𝑖 )3 = βˆ’(𝟏 + π’Š )πŸ‘ Using (a + b) 3 = a3 + b3 + 3ab(a + b) = βˆ’(13 + 𝑖3 + 3 Γ— 1 Γ— 𝑖 (1 + 𝑖)) = βˆ’(1 + π’ŠπŸ‘ +3𝑖 (1 + 𝑖)) = βˆ’(1 + π’ŠπŸ Γ— π’Š +3𝑖 (1 + 𝑖)) Putting i2 = βˆ’πŸ = βˆ’(1 +(βˆ’πŸ) Γ— 𝑖 +3𝑖 (1 + 𝑖)) = βˆ’(1 βˆ’π‘– +3𝑖 (1 + 𝑖)) = βˆ’(1 βˆ’π‘– +3𝑖+3𝑖 Γ— 𝑖) = βˆ’(1+2𝑖+3π’Š^𝟐 ) Putting i2 = βˆ’πŸ = βˆ’(1+2𝑖+3 Γ— βˆ’πŸ) = βˆ’(1+2π‘–βˆ’3) = βˆ’(2π‘–βˆ’2) = βˆ’2𝑖+2 = πŸβˆ’πŸπ’Š

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.