{"id":8263,"date":"2021-11-20T00:38:41","date_gmt":"2021-11-19T19:08:41","guid":{"rendered":"https:\/\/mathemerize.com\/?p=8263"},"modified":"2021-11-20T01:29:08","modified_gmt":"2021-11-19T19:59:08","slug":"by-using-binomial-theorem-expand-1-x-x23","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/by-using-binomial-theorem-expand-1-x-x23\/","title":{"rendered":"By using binomial theorem, expand \\((1 + x + x^2)^3\\)."},"content":{"rendered":"

Solution :<\/h2>\n

Let y = x + \\(x^2\\). Then,<\/p>\n

\\((1 + x + x^2)^3\\) = \\((1 + y)^3\\)<\/p>\n

= \\(^3C_0\\) + \\(^3C_1 y\\) + \\(^3C_2 y^2\\) + \\(^3C_3 y^3\\)<\/p>\n

= \\(1 + 3y + 3y^2 + y^3\\) = 1 + 3\\((x + x^2)\\) + 3\\((x + x^2)^2\\) + \\((x + x^2)^3\\)<\/p>\n

= \\(x^6 + 3x^5 + 6x^4 + 7x^3 + 6x^2 + 3x + 1\\)<\/p>\n

\n

Similar Questions<\/h3>\n

Find the middle term in the expansion of \\(({2\\over 3}x^2 \u2013 {3\\over 2x})^{20}\\).<\/a><\/p>\n

Find the 9th term in the expansion of \\(({x\\over a} \u2013 {3a\\over x^2})^{12}\\).<\/a><\/p>\n

Find the 10th term in the binomial expansion of \\((2x^2 + {1\\over x})^{12}\\).<\/a><\/p>\n

Which is larger \\((1.01)^{1000000}\\) or 10,000?<\/a><\/p>\n

Find the middle term in the expansion of \\((3x \u2013 {x^3\\over 6})^7\\).<\/a><\/p>\n","protected":false},"excerpt":{"rendered":"

Solution : Let y = x + \\(x^2\\). Then, \\((1 + x + x^2)^3\\) = \\((1 + y)^3\\) = \\(^3C_0\\) + \\(^3C_1 y\\) + \\(^3C_2 y^2\\) + \\(^3C_3 y^3\\) = \\(1 + 3y + 3y^2 + y^3\\) = 1 + 3\\((x + x^2)\\) + 3\\((x + x^2)^2\\) + \\((x + x^2)^3\\) = \\(x^6 + 3x^5 …<\/p>\n

By using binomial theorem, expand \\((1 + x + x^2)^3\\).<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[177,43],"tags":[],"yoast_head":"\nBy using binomial theorem, expand \\((1 + x + x^2)^3\\).<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/by-using-binomial-theorem-expand-1-x-x23\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"By using binomial theorem, expand \\((1 + x + x^2)^3\\).\" \/>\n<meta property=\"og:description\" content=\"Solution : Let y = x + (x^2). Then, ((1 + x + x^2)^3) = ((1 + y)^3) = (^3C_0) + (^3C_1 y) + (^3C_2 y^2) + (^3C_3 y^3) = (1 + 3y + 3y^2 + y^3) = 1 + 3((x + x^2)) + 3((x + x^2)^2) + ((x + x^2)^3) = (x^6 + 3x^5 … By using binomial theorem, expand ((1 + x + x^2)^3). 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Then, ((1 + x + x^2)^3) = ((1 + y)^3) = (^3C_0) + (^3C_1 y) + (^3C_2 y^2) + (^3C_3 y^3) = (1 + 3y + 3y^2 + y^3) = 1 + 3((x + x^2)) + 3((x + x^2)^2) + ((x + x^2)^3) = (x^6 + 3x^5 … By using binomial theorem, expand ((1 + x + x^2)^3). 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