[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]Understand the problem
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[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]We have to use algebraic expression and expansion e.g- \(( x + 1) ^{2}\) = \(x^2\) + 2x + 1 ; \(( x +1)^{3}\) = \(x^{3}\) + 3\(x^2\) +3x + 1
Can you think about this kind of case a bit more ?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]Note that \(x^3\)+ 1 = (\(x+1\))(\(x^{2} - x + 1 \))
So, \(\alpha\) is a root of \(x^{2} - x + 1\) we have (\(\alpha^3\) + 1) = (\(\alpha+1\))(\(\alpha^{2}\) - \(\alpha\) + 1 ) = ( \(\alpha\) + 1 ) = 0
Can we think along this line ?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]\(\alpha^{3}\) = -1
||ly \(\beta^{3}\) = -1 [ when \(\beta\) is another root (\(x^{2} - x + 1 \)) ]
So, (\(x^{2} - x + 1 \)) = ( x - \(\alpha\) )( x - \(\beta\)) = \(x^{2}\) - (\(\alpha + \beta\))x + \(\alpha\)\(\beta\).
Can you get the expression of \(\beta\) in terms of \(\alpha\) ?
Is the cloud getting clear now ?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]\(\alpha\) + \(\beta\) = 1
\(\alpha\)\(\beta\) = 1 implying \(\beta\) = 1/ \(\alpha\)
So, \(\alpha^{2018}\) + \(\alpha^{-2018}\) = \(\alpha^{2018}\) + \(\beta^{2018}\)
Note that : \(\alpha^{2}\) + \(\beta^{2}\) = \(\alpha + \beta )^{2}\) -2\(\alpha\)\(\beta\) = 1 - 2\(\alpha\)\(\beta\) = 1 -2 = -1
\(\alpha^{2018}\) + \(\beta^{2018}\) = \(\alpha^{3x672}\) . \(\alpha^{2}\) + \(\beta^{3x672}\) . \(\beta^{2}\)
= (\(\alpha^{3})^{672}\).\(\alpha^{2}\) + (\(\beta^{3})^{672}\).\(\beta^{2}\)
= (\(-1^{672}\)).\(\alpha^{2}\) + (\(-1^{672}\)).\(\beta^{2}\)
= \(\alpha^{2}\) + \(\beta^{2}\)
= -1
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[/et_pb_text][et_pb_code _builder_version="3.26.4" _i="4" _address="0.1.0.4"][/et_pb_code][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" min_height="12px" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="5" _address="0.1.0.5"]Connected Program at Cheenta
[/et_pb_text][et_pb_blurb title="College Mathematics Program" url="https://www.cheenta.com/collegeprogram/" image="https://www.cheenta.com/wp-content/uploads/2018/03/College-1.png" _builder_version="3.23.3" header_font="||||||||" header_text_color="#e02b20" header_font_size="48px" border_color_all="#e02b20" link_option_url="https://www.cheenta.com/collegeprogram/" _i="6" _address="0.1.0.6"]The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]
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[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first![/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]We have to use algebraic expression and expansion e.g- \(( x + 1) ^{2}\) = \(x^2\) + 2x + 1 ; \(( x +1)^{3}\) = \(x^{3}\) + 3\(x^2\) +3x + 1
Can you think about this kind of case a bit more ?[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]Note that \(x^3\)+ 1 = (\(x+1\))(\(x^{2} - x + 1 \))
So, \(\alpha\) is a root of \(x^{2} - x + 1\) we have (\(\alpha^3\) + 1) = (\(\alpha+1\))(\(\alpha^{2}\) - \(\alpha\) + 1 ) = ( \(\alpha\) + 1 ) = 0
Can we think along this line ?[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]\(\alpha^{3}\) = -1
||ly \(\beta^{3}\) = -1 [ when \(\beta\) is another root (\(x^{2} - x + 1 \)) ]
So, (\(x^{2} - x + 1 \)) = ( x - \(\alpha\) )( x - \(\beta\)) = \(x^{2}\) - (\(\alpha + \beta\))x + \(\alpha\)\(\beta\).
Can you get the expression of \(\beta\) in terms of \(\alpha\) ?
Is the cloud getting clear now ?[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]\(\alpha\) + \(\beta\) = 1
\(\alpha\)\(\beta\) = 1 implying \(\beta\) = 1/ \(\alpha\)
So, \(\alpha^{2018}\) + \(\alpha^{-2018}\) = \(\alpha^{2018}\) + \(\beta^{2018}\)
Note that : \(\alpha^{2}\) + \(\beta^{2}\) = \(\alpha + \beta )^{2}\) -2\(\alpha\)\(\beta\) = 1 - 2\(\alpha\)\(\beta\) = 1 -2 = -1
\(\alpha^{2018}\) + \(\beta^{2018}\) = \(\alpha^{3x672}\) . \(\alpha^{2}\) + \(\beta^{3x672}\) . \(\beta^{2}\)
= (\(\alpha^{3})^{672}\).\(\alpha^{2}\) + (\(\beta^{3})^{672}\).\(\beta^{2}\)
= (\(-1^{672}\)).\(\alpha^{2}\) + (\(-1^{672}\)).\(\beta^{2}\)
= \(\alpha^{2}\) + \(\beta^{2}\)
= -1
[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]Watch the video
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it can be solved more easily.
Hi,
Yes, you are right but this method came into my mind first so I typed it down :). You can write your solution so that the readers can see that as well. Thanks a lot for seeing this. Keep reading and don't forget to comment. Your valuable comments/appreciations inspire us to create more contents.
Regards,
Arnab